Revisions to Integral involving InverseErf: closed-form expression and asymptotics

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edited Oct 17, 2017 at 17:01

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Answer from comments.

Let $F(0)=\text{erf}\left(\sqrt{t}\right)^4$$F(t)=\text{erf}\left(\sqrt{t}\right)^4$.Then $F'(t)=\frac{4 e^{-t} \text{erf}\left(\sqrt{t}\right)^3}{\sqrt{\pi } \sqrt{t}}$ and $F(0)=0$. $$\mathcal{L}_t\left[F'(t)\right](s)=s \left(\mathcal{L}_t[F(t)](s)\right)-F(0)$$ $$\mathcal{L}_t\left[\frac{4 e^{-t} \text{erf}\left(\sqrt{t}\right)^3}{\sqrt{\pi } \sqrt{t}}\right](s)=s \left(\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)\right)-0$$

$$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)=\frac{4 \left(\mathcal{L}_t\left[\frac{\text{erf}\left(\sqrt{t}\right)^3}{\sqrt{t}}\right](1+s)\right)}{\sqrt{\pi } s}$$ Solve Laplace Transform with integral:

$$\int_0^{\infty } \frac{\text{erf}\left(\sqrt{t}\right)^3 \exp (-(s+1) t)}{\sqrt{t}} \, dt$$ Integrating with parts and substituting Sqrt[t]=x

$${\frac {\sqrt {\pi}}{\sqrt {s+1}}}-\int_{0}^{\infty }\!6\,{\frac { {\rm erf} \left(\sqrt {s+1}x\right) \left( {\rm erf} \left(x\right) \right) ^{2}{{\rm e}^{-{x}^{2}}}}{\sqrt {s+1}}}\,{\rm d}x $$

Consider the parametric integral : $$ I(a,b) = \int_{0}^{\infty}\!\!\!\text{erf}(ax)\,\text{erf}(bx)\,\text{erf}(bx)\,e^{-x^2}dx \tag{1}$$ and notice that:

$$\frac{\partial }{\partial a}\frac{\partial I(a,b)}{\partial b}=\int_0^{\infty } \frac{8 e^{-x^2-a^2 x^2-b^2 x^2} x^2 \text{erf}(b x)}{\pi } \, dx=\frac{4 b}{\pi ^{3/2} \left(a^2+b^2+1\right) \left(a^2+2 b^2+1\right)}+\frac{4 \tan ^{-1}\left(\frac{b}{\sqrt{a^2+b^2+1}}\right)}{\pi ^{3/2} \left(a^2+b^2+1\right)^{3/2}}$$ integrating over $b\in \{0,1\}$ and $a\in \{0,\sqrt{s+1}\}$

 Integrate[(4 b)/((1 + a^2 + b^2) (1 + a^2 + 2 b^2) \[Pi]^(3/2)) + (
 4 ArcTan[b/Sqrt[1 + a^2 + b^2]])/((1 + a^2 + b^2)^(3/2) \[Pi]^(
 /2)), {b, 0, 1}, Assumptions -> a > 0]
 (*(4 ArcTan[1/Sqrt[2 + a^2]])/((1 + a^2) Sqrt[2 + a^2] \[Pi]^(3/2))*)

 Integrate[(4 ArcTan[1/Sqrt[2 + a^2]])/((1 + a^2) Sqrt[2 + a^2] \[Pi]^(3/2)), {a,
 0, Sqrt[s + 1]}]
 (*----*)

$$\int_0^{\sqrt{s+1}} \frac{4 \tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2} \pi ^{3/2}} \, da$$ Connect everything together:

$$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)=\frac{4 \sqrt{\frac{1}{s+1}}}{s}-96\frac{\int_0^{\sqrt{s+1}} \frac{\tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2}} \, da}{\pi ^2 s \sqrt{s+1}}$$

.EDIT 15.02.2017.

Finding Integral:

$\int_0^{\sqrt{s+1}} \frac{\tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2}} \, da=\frac{1}{4} \pi \tan ^{-1}\left(\sqrt{s+1}\right)-\int_0^1 \frac{\tan ^{-1}\left(\frac{\sqrt{s+1}}{\sqrt{a^2+2}}\right)}{\sqrt{a^2+2} \left(a^2+1\right)} \, da$

$\int_0^1 \frac{\tan ^{-1}\left(\frac{\sqrt{s+1}}{\sqrt{a^2+2}}\right)}{\sqrt{a^2+2} \left(a^2+1\right)} \, da = \frac{1}{6} \left(3 \sqrt{s+1} X(s)+\pi \tan ^{-1}\left(\frac{\sqrt{s+1}}{\sqrt{3}}\right)\right) $

where X[s]is:

$X(s)=\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}+\sqrt{s+1} \sin (x)\right)} \, dx+\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}-\sqrt{s+1} \sin (x)\right)} \, dx$

Check numerics:

NIntegrate[ArcTan[Sqrt[s + 1]/Sqrt[x^2 + 2]]/(Sqrt[x^2 + 2]*(x^2 + 1)) /. 
s -> 1, {x, 0, 1}]
(* 0.396282 *)

X[s_?NumericQ] := 
NIntegrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
I Sqrt[2] + Sqrt[s + 1]* Sin[x]), {x, ArcSin[Sqrt[3]*Sqrt[2]/3], 
Pi/2}] + 
NIntegrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
I Sqrt[2] - Sqrt[s + 1]* Sin[x]), {x, ArcSin[Sqrt[3]*Sqrt[2]/3], 
Pi/2}]
1/6 (\[Pi] ArcTan[Sqrt[1 + s]/Sqrt[3]] + 3 Sqrt[1 + s] X[s]) /. 
s -> 1 // N
(* 0.396282 *)
Ok works.

Finding a integral :

$\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}-\sqrt{s+1} \sin (x)\right)} \, dx=?$

MMA has problem to caluculate this integral with limits,but without limits is: where a is Sqrt[s+1]

 Integrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
 I Sqrt[2] -a*Sin[x]), x]
 (*the result is a very long*)

Answer from comments.

Let $F(0)=\text{erf}\left(\sqrt{t}\right)^4$.Then $F'(t)=\frac{4 e^{-t} \text{erf}\left(\sqrt{t}\right)^3}{\sqrt{\pi } \sqrt{t}}$ and $F(0)=0$. $$\mathcal{L}_t\left[F'(t)\right](s)=s \left(\mathcal{L}_t[F(t)](s)\right)-F(0)$$ $$\mathcal{L}_t\left[\frac{4 e^{-t} \text{erf}\left(\sqrt{t}\right)^3}{\sqrt{\pi } \sqrt{t}}\right](s)=s \left(\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)\right)-0$$

$$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)=\frac{4 \left(\mathcal{L}_t\left[\frac{\text{erf}\left(\sqrt{t}\right)^3}{\sqrt{t}}\right](1+s)\right)}{\sqrt{\pi } s}$$ Solve Laplace Transform with integral:

$$\int_0^{\infty } \frac{\text{erf}\left(\sqrt{t}\right)^3 \exp (-(s+1) t)}{\sqrt{t}} \, dt$$ Integrating with parts and substituting Sqrt[t]=x

$${\frac {\sqrt {\pi}}{\sqrt {s+1}}}-\int_{0}^{\infty }\!6\,{\frac { {\rm erf} \left(\sqrt {s+1}x\right) \left( {\rm erf} \left(x\right) \right) ^{2}{{\rm e}^{-{x}^{2}}}}{\sqrt {s+1}}}\,{\rm d}x $$

Consider the parametric integral : $$ I(a,b) = \int_{0}^{\infty}\!\!\!\text{erf}(ax)\,\text{erf}(bx)\,\text{erf}(bx)\,e^{-x^2}dx \tag{1}$$ and notice that:

$$\frac{\partial }{\partial a}\frac{\partial I(a,b)}{\partial b}=\int_0^{\infty } \frac{8 e^{-x^2-a^2 x^2-b^2 x^2} x^2 \text{erf}(b x)}{\pi } \, dx=\frac{4 b}{\pi ^{3/2} \left(a^2+b^2+1\right) \left(a^2+2 b^2+1\right)}+\frac{4 \tan ^{-1}\left(\frac{b}{\sqrt{a^2+b^2+1}}\right)}{\pi ^{3/2} \left(a^2+b^2+1\right)^{3/2}}$$ integrating over $b\in \{0,1\}$ and $a\in \{0,\sqrt{s+1}\}$

 Integrate[(4 b)/((1 + a^2 + b^2) (1 + a^2 + 2 b^2) \[Pi]^(3/2)) + (
 4 ArcTan[b/Sqrt[1 + a^2 + b^2]])/((1 + a^2 + b^2)^(3/2) \[Pi]^(
 /2)), {b, 0, 1}, Assumptions -> a > 0]
 (*(4 ArcTan[1/Sqrt[2 + a^2]])/((1 + a^2) Sqrt[2 + a^2] \[Pi]^(3/2))*)

 Integrate[(4 ArcTan[1/Sqrt[2 + a^2]])/((1 + a^2) Sqrt[2 + a^2] \[Pi]^(3/2)), {a,
 0, Sqrt[s + 1]}]
 (*----*)

$$\int_0^{\sqrt{s+1}} \frac{4 \tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2} \pi ^{3/2}} \, da$$ Connect everything together:

$$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)=\frac{4 \sqrt{\frac{1}{s+1}}}{s}-96\frac{\int_0^{\sqrt{s+1}} \frac{\tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2}} \, da}{\pi ^2 s \sqrt{s+1}}$$

.EDIT 15.02.2017.

Finding Integral:

$\int_0^{\sqrt{s+1}} \frac{\tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2}} \, da=\frac{1}{4} \pi \tan ^{-1}\left(\sqrt{s+1}\right)-\int_0^1 \frac{\tan ^{-1}\left(\frac{\sqrt{s+1}}{\sqrt{a^2+2}}\right)}{\sqrt{a^2+2} \left(a^2+1\right)} \, da$

$\int_0^1 \frac{\tan ^{-1}\left(\frac{\sqrt{s+1}}{\sqrt{a^2+2}}\right)}{\sqrt{a^2+2} \left(a^2+1\right)} \, da = \frac{1}{6} \left(3 \sqrt{s+1} X(s)+\pi \tan ^{-1}\left(\frac{\sqrt{s+1}}{\sqrt{3}}\right)\right) $

where X[s]is:

$X(s)=\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}+\sqrt{s+1} \sin (x)\right)} \, dx+\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}-\sqrt{s+1} \sin (x)\right)} \, dx$

Check numerics:

NIntegrate[ArcTan[Sqrt[s + 1]/Sqrt[x^2 + 2]]/(Sqrt[x^2 + 2]*(x^2 + 1)) /. 
s -> 1, {x, 0, 1}]
(* 0.396282 *)

X[s_?NumericQ] := 
NIntegrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
I Sqrt[2] + Sqrt[s + 1]* Sin[x]), {x, ArcSin[Sqrt[3]*Sqrt[2]/3], 
Pi/2}] + 
NIntegrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
I Sqrt[2] - Sqrt[s + 1]* Sin[x]), {x, ArcSin[Sqrt[3]*Sqrt[2]/3], 
Pi/2}]
1/6 (\[Pi] ArcTan[Sqrt[1 + s]/Sqrt[3]] + 3 Sqrt[1 + s] X[s]) /. 
s -> 1 // N
(* 0.396282 *)
Ok works.

Finding a integral :

$\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}-\sqrt{s+1} \sin (x)\right)} \, dx=?$

MMA has problem to caluculate this integral with limits,but without limits is: where a is Sqrt[s+1]

 Integrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
 I Sqrt[2] -a*Sin[x]), x]
 (*the result is a very long*)

Answer from comments.

Let $F(t)=\text{erf}\left(\sqrt{t}\right)^4$.Then $F'(t)=\frac{4 e^{-t} \text{erf}\left(\sqrt{t}\right)^3}{\sqrt{\pi } \sqrt{t}}$ and $F(0)=0$. $$\mathcal{L}_t\left[F'(t)\right](s)=s \left(\mathcal{L}_t[F(t)](s)\right)-F(0)$$ $$\mathcal{L}_t\left[\frac{4 e^{-t} \text{erf}\left(\sqrt{t}\right)^3}{\sqrt{\pi } \sqrt{t}}\right](s)=s \left(\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)\right)-0$$

$$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)=\frac{4 \left(\mathcal{L}_t\left[\frac{\text{erf}\left(\sqrt{t}\right)^3}{\sqrt{t}}\right](1+s)\right)}{\sqrt{\pi } s}$$ Solve Laplace Transform with integral:

$$\int_0^{\infty } \frac{\text{erf}\left(\sqrt{t}\right)^3 \exp (-(s+1) t)}{\sqrt{t}} \, dt$$ Integrating with parts and substituting Sqrt[t]=x

$${\frac {\sqrt {\pi}}{\sqrt {s+1}}}-\int_{0}^{\infty }\!6\,{\frac { {\rm erf} \left(\sqrt {s+1}x\right) \left( {\rm erf} \left(x\right) \right) ^{2}{{\rm e}^{-{x}^{2}}}}{\sqrt {s+1}}}\,{\rm d}x $$

Consider the parametric integral : $$ I(a,b) = \int_{0}^{\infty}\!\!\!\text{erf}(ax)\,\text{erf}(bx)\,\text{erf}(bx)\,e^{-x^2}dx \tag{1}$$ and notice that:

$$\frac{\partial }{\partial a}\frac{\partial I(a,b)}{\partial b}=\int_0^{\infty } \frac{8 e^{-x^2-a^2 x^2-b^2 x^2} x^2 \text{erf}(b x)}{\pi } \, dx=\frac{4 b}{\pi ^{3/2} \left(a^2+b^2+1\right) \left(a^2+2 b^2+1\right)}+\frac{4 \tan ^{-1}\left(\frac{b}{\sqrt{a^2+b^2+1}}\right)}{\pi ^{3/2} \left(a^2+b^2+1\right)^{3/2}}$$ integrating over $b\in \{0,1\}$ and $a\in \{0,\sqrt{s+1}\}$

 Integrate[(4 b)/((1 + a^2 + b^2) (1 + a^2 + 2 b^2) \[Pi]^(3/2)) + (
 4 ArcTan[b/Sqrt[1 + a^2 + b^2]])/((1 + a^2 + b^2)^(3/2) \[Pi]^(
 /2)), {b, 0, 1}, Assumptions -> a > 0]
 (*(4 ArcTan[1/Sqrt[2 + a^2]])/((1 + a^2) Sqrt[2 + a^2] \[Pi]^(3/2))*)

 Integrate[(4 ArcTan[1/Sqrt[2 + a^2]])/((1 + a^2) Sqrt[2 + a^2] \[Pi]^(3/2)), {a,
 0, Sqrt[s + 1]}]
 (*----*)

$$\int_0^{\sqrt{s+1}} \frac{4 \tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2} \pi ^{3/2}} \, da$$ Connect everything together:

$$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)=\frac{4 \sqrt{\frac{1}{s+1}}}{s}-96\frac{\int_0^{\sqrt{s+1}} \frac{\tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2}} \, da}{\pi ^2 s \sqrt{s+1}}$$

.EDIT 15.02.2017.

Finding Integral:

$\int_0^{\sqrt{s+1}} \frac{\tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2}} \, da=\frac{1}{4} \pi \tan ^{-1}\left(\sqrt{s+1}\right)-\int_0^1 \frac{\tan ^{-1}\left(\frac{\sqrt{s+1}}{\sqrt{a^2+2}}\right)}{\sqrt{a^2+2} \left(a^2+1\right)} \, da$

$\int_0^1 \frac{\tan ^{-1}\left(\frac{\sqrt{s+1}}{\sqrt{a^2+2}}\right)}{\sqrt{a^2+2} \left(a^2+1\right)} \, da = \frac{1}{6} \left(3 \sqrt{s+1} X(s)+\pi \tan ^{-1}\left(\frac{\sqrt{s+1}}{\sqrt{3}}\right)\right) $

where X[s]is:

$X(s)=\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}+\sqrt{s+1} \sin (x)\right)} \, dx+\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}-\sqrt{s+1} \sin (x)\right)} \, dx$

Check numerics:

NIntegrate[ArcTan[Sqrt[s + 1]/Sqrt[x^2 + 2]]/(Sqrt[x^2 + 2]*(x^2 + 1)) /. 
s -> 1, {x, 0, 1}]
(* 0.396282 *)

X[s_?NumericQ] := 
NIntegrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
I Sqrt[2] + Sqrt[s + 1]* Sin[x]), {x, ArcSin[Sqrt[3]*Sqrt[2]/3], 
Pi/2}] + 
NIntegrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
I Sqrt[2] - Sqrt[s + 1]* Sin[x]), {x, ArcSin[Sqrt[3]*Sqrt[2]/3], 
Pi/2}]
1/6 (\[Pi] ArcTan[Sqrt[1 + s]/Sqrt[3]] + 3 Sqrt[1 + s] X[s]) /. 
s -> 1 // N
(* 0.396282 *)
Ok works.

Finding a integral :

$\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}-\sqrt{s+1} \sin (x)\right)} \, dx=?$

MMA has problem to caluculate this integral with limits,but without limits is: where a is Sqrt[s+1]

 Integrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
 I Sqrt[2] -a*Sin[x]), x]
 (*the result is a very long*)

added 268 characters in body

Source Link

edited Feb 15, 2017 at 21:41

Mariusz Iwaniuk

16.4k
1
27
46

Answer from comments.

Let $F(0)=\text{erf}\left(\sqrt{t}\right)^4$.Then $F'(t)=\frac{4 e^{-t} \text{erf}\left(\sqrt{t}\right)^3}{\sqrt{\pi } \sqrt{t}}$ and $F(0)=0$. $$\mathcal{L}_t\left[F'(t)\right](s)=s \left(\mathcal{L}_t[F(t)](s)\right)-F(0)$$ $$\mathcal{L}_t\left[\frac{4 e^{-t} \text{erf}\left(\sqrt{t}\right)^3}{\sqrt{\pi } \sqrt{t}}\right](s)=s \left(\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)\right)-0$$

$$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)=\frac{4 \left(\mathcal{L}_t\left[\frac{\text{erf}\left(\sqrt{t}\right)^3}{\sqrt{t}}\right](1+s)\right)}{\sqrt{\pi } s}$$ Solve Laplace Transform with integral:

$$\int_0^{\infty } \frac{\text{erf}\left(\sqrt{t}\right)^3 \exp (-(s+1) t)}{\sqrt{t}} \, dt$$ Integrating with parts and substituting Sqrt[t]=x

$${\frac {\sqrt {\pi}}{\sqrt {s+1}}}-\int_{0}^{\infty }\!6\,{\frac { {\rm erf} \left(\sqrt {s+1}x\right) \left( {\rm erf} \left(x\right) \right) ^{2}{{\rm e}^{-{x}^{2}}}}{\sqrt {s+1}}}\,{\rm d}x $$

Consider the parametric integral : $$ I(a,b) = \int_{0}^{\infty}\!\!\!\text{erf}(ax)\,\text{erf}(bx)\,\text{erf}(bx)\,e^{-x^2}dx \tag{1}$$ and notice that:

$$\frac{\partial }{\partial a}\frac{\partial I(a,b)}{\partial b}=\int_0^{\infty } \frac{8 e^{-x^2-a^2 x^2-b^2 x^2} x^2 \text{erf}(b x)}{\pi } \, dx=\frac{4 b}{\pi ^{3/2} \left(a^2+b^2+1\right) \left(a^2+2 b^2+1\right)}+\frac{4 \tan ^{-1}\left(\frac{b}{\sqrt{a^2+b^2+1}}\right)}{\pi ^{3/2} \left(a^2+b^2+1\right)^{3/2}}$$ integrating over $b\in \{0,1\}$ and $a\in \{0,\sqrt{s+1}\}$

 Integrate[(4 b)/((1 + a^2 + b^2) (1 + a^2 + 2 b^2) \[Pi]^(3/2)) + (
 4 ArcTan[b/Sqrt[1 + a^2 + b^2]])/((1 + a^2 + b^2)^(3/2) \[Pi]^(
 /2)), {b, 0, 1}, Assumptions -> a > 0]
 (*(4 ArcTan[1/Sqrt[2 + a^2]])/((1 + a^2) Sqrt[2 + a^2] \[Pi]^(3/2))*)

 Integrate[(4 ArcTan[1/Sqrt[2 + a^2]])/((1 + a^2) Sqrt[2 + a^2] \[Pi]^(3/2)), {a,
 0, Sqrt[s + 1]}]
 (*----*)

$$\int_0^{\sqrt{s+1}} \frac{4 \tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2} \pi ^{3/2}} \, da$$ Connect everything together:

$$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)=\frac{4 \sqrt{\frac{1}{s+1}}}{s}-96\frac{\int_0^{\sqrt{s+1}} \frac{\tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2}} \, da}{\pi ^2 s \sqrt{s+1}}$$

.EDIT 15.02.2017.

$\int_{0}^{1}\!{\frac {1}{ \left( {y}^{2}+1 \right) \sqrt {{y}^{2}+2}} \arctan \left( {\frac {\sqrt {s+1}}{\sqrt {{y}^{2}+2}}} \right) } \,{\rm d}y = \frac{1}{6} \left(3 \sqrt{s+1} X(s)+\pi \tan ^{-1}\left(\frac{\sqrt{s+1}}{\sqrt{3}}\right)\right) $ Finding Integral:

$\int_0^{\sqrt{s+1}} \frac{\tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2}} \, da=\frac{1}{4} \pi \tan ^{-1}\left(\sqrt{s+1}\right)-\int_0^1 \frac{\tan ^{-1}\left(\frac{\sqrt{s+1}}{\sqrt{a^2+2}}\right)}{\sqrt{a^2+2} \left(a^2+1\right)} \, da$

$\int_0^1 \frac{\tan ^{-1}\left(\frac{\sqrt{s+1}}{\sqrt{a^2+2}}\right)}{\sqrt{a^2+2} \left(a^2+1\right)} \, da = \frac{1}{6} \left(3 \sqrt{s+1} X(s)+\pi \tan ^{-1}\left(\frac{\sqrt{s+1}}{\sqrt{3}}\right)\right) $

where X[s]is:

$X(s)=\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}+\sqrt{s+1} \sin (x)\right)} \, dx+\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}-\sqrt{s+1} \sin (x)\right)} \, dx$

Check numerics:

NIntegrate[ArcTan[Sqrt[s + 1]/Sqrt[x^2 + 2]]/(Sqrt[x^2 + 2]*(x^2 + 1)) /. 
s -> 1, {x, 0, 1}]
(* 0.396282 *)

X[s_?NumericQ] := 
NIntegrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
I Sqrt[2] + Sqrt[s + 1]* Sin[x]), {x, ArcSin[Sqrt[3]*Sqrt[2]/3], 
Pi/2}] + 
NIntegrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
I Sqrt[2] - Sqrt[s + 1]* Sin[x]), {x, ArcSin[Sqrt[3]*Sqrt[2]/3], 
Pi/2}]
1/6 (\[Pi] ArcTan[Sqrt[1 + s]/Sqrt[3]] + 3 Sqrt[1 + s] X[s]) /. 
s -> 1 // N
(* 0.396282 *)
Ok works.

Finding a integral :

$\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}-\sqrt{s+1} \sin (x)\right)} \, dx=?$

MMA has problem to caluculate this integral with limits,but without limits is: where a is Sqrt[s+1]

 Integrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
 I Sqrt[2] -a*Sin[x]), x]
 (*the result is a very long*)

Answer from comments.

Let $F(0)=\text{erf}\left(\sqrt{t}\right)^4$.Then $F'(t)=\frac{4 e^{-t} \text{erf}\left(\sqrt{t}\right)^3}{\sqrt{\pi } \sqrt{t}}$ and $F(0)=0$. $$\mathcal{L}_t\left[F'(t)\right](s)=s \left(\mathcal{L}_t[F(t)](s)\right)-F(0)$$ $$\mathcal{L}_t\left[\frac{4 e^{-t} \text{erf}\left(\sqrt{t}\right)^3}{\sqrt{\pi } \sqrt{t}}\right](s)=s \left(\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)\right)-0$$

$$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)=\frac{4 \left(\mathcal{L}_t\left[\frac{\text{erf}\left(\sqrt{t}\right)^3}{\sqrt{t}}\right](1+s)\right)}{\sqrt{\pi } s}$$ Solve Laplace Transform with integral:

$$\int_0^{\infty } \frac{\text{erf}\left(\sqrt{t}\right)^3 \exp (-(s+1) t)}{\sqrt{t}} \, dt$$ Integrating with parts and substituting Sqrt[t]=x

$${\frac {\sqrt {\pi}}{\sqrt {s+1}}}-\int_{0}^{\infty }\!6\,{\frac { {\rm erf} \left(\sqrt {s+1}x\right) \left( {\rm erf} \left(x\right) \right) ^{2}{{\rm e}^{-{x}^{2}}}}{\sqrt {s+1}}}\,{\rm d}x $$

Consider the parametric integral : $$ I(a,b) = \int_{0}^{\infty}\!\!\!\text{erf}(ax)\,\text{erf}(bx)\,\text{erf}(bx)\,e^{-x^2}dx \tag{1}$$ and notice that:

$$\frac{\partial }{\partial a}\frac{\partial I(a,b)}{\partial b}=\int_0^{\infty } \frac{8 e^{-x^2-a^2 x^2-b^2 x^2} x^2 \text{erf}(b x)}{\pi } \, dx=\frac{4 b}{\pi ^{3/2} \left(a^2+b^2+1\right) \left(a^2+2 b^2+1\right)}+\frac{4 \tan ^{-1}\left(\frac{b}{\sqrt{a^2+b^2+1}}\right)}{\pi ^{3/2} \left(a^2+b^2+1\right)^{3/2}}$$ integrating over $b\in \{0,1\}$ and $a\in \{0,\sqrt{s+1}\}$

 Integrate[(4 b)/((1 + a^2 + b^2) (1 + a^2 + 2 b^2) \[Pi]^(3/2)) + (
 4 ArcTan[b/Sqrt[1 + a^2 + b^2]])/((1 + a^2 + b^2)^(3/2) \[Pi]^(
 /2)), {b, 0, 1}, Assumptions -> a > 0]
 (*(4 ArcTan[1/Sqrt[2 + a^2]])/((1 + a^2) Sqrt[2 + a^2] \[Pi]^(3/2))*)

 Integrate[(4 ArcTan[1/Sqrt[2 + a^2]])/((1 + a^2) Sqrt[2 + a^2] \[Pi]^(3/2)), {a,
 0, Sqrt[s + 1]}]
 (*----*)

$$\int_0^{\sqrt{s+1}} \frac{4 \tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2} \pi ^{3/2}} \, da$$ Connect everything together:

$$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)=\frac{4 \sqrt{\frac{1}{s+1}}}{s}-96\frac{\int_0^{\sqrt{s+1}} \frac{\tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2}} \, da}{\pi ^2 s \sqrt{s+1}}$$

.EDIT 15.02.2017.

$\int_{0}^{1}\!{\frac {1}{ \left( {y}^{2}+1 \right) \sqrt {{y}^{2}+2}} \arctan \left( {\frac {\sqrt {s+1}}{\sqrt {{y}^{2}+2}}} \right) } \,{\rm d}y = \frac{1}{6} \left(3 \sqrt{s+1} X(s)+\pi \tan ^{-1}\left(\frac{\sqrt{s+1}}{\sqrt{3}}\right)\right) $

where X[s]is:

$X(s)=\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}+\sqrt{s+1} \sin (x)\right)} \, dx+\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}-\sqrt{s+1} \sin (x)\right)} \, dx$

Check numerics:

NIntegrate[ArcTan[Sqrt[s + 1]/Sqrt[x^2 + 2]]/(Sqrt[x^2 + 2]*(x^2 + 1)) /. 
s -> 1, {x, 0, 1}]
(* 0.396282 *)

X[s_?NumericQ] := 
NIntegrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
I Sqrt[2] + Sqrt[s + 1]* Sin[x]), {x, ArcSin[Sqrt[3]*Sqrt[2]/3], 
Pi/2}] + 
NIntegrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
I Sqrt[2] - Sqrt[s + 1]* Sin[x]), {x, ArcSin[Sqrt[3]*Sqrt[2]/3], 
Pi/2}]
1/6 (\[Pi] ArcTan[Sqrt[1 + s]/Sqrt[3]] + 3 Sqrt[1 + s] X[s]) /. 
s -> 1 // N
(* 0.396282 *)
Ok works.

Finding a integral :

$\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}-\sqrt{s+1} \sin (x)\right)} \, dx=?$

MMA has problem to caluculate this integral with limits,but without limits is: where a is Sqrt[s+1]

 Integrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
 I Sqrt[2] -a*Sin[x]), x]
 (*the result is a very long*)

Answer from comments.

Let $F(0)=\text{erf}\left(\sqrt{t}\right)^4$.Then $F'(t)=\frac{4 e^{-t} \text{erf}\left(\sqrt{t}\right)^3}{\sqrt{\pi } \sqrt{t}}$ and $F(0)=0$. $$\mathcal{L}_t\left[F'(t)\right](s)=s \left(\mathcal{L}_t[F(t)](s)\right)-F(0)$$ $$\mathcal{L}_t\left[\frac{4 e^{-t} \text{erf}\left(\sqrt{t}\right)^3}{\sqrt{\pi } \sqrt{t}}\right](s)=s \left(\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)\right)-0$$

$$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)=\frac{4 \left(\mathcal{L}_t\left[\frac{\text{erf}\left(\sqrt{t}\right)^3}{\sqrt{t}}\right](1+s)\right)}{\sqrt{\pi } s}$$ Solve Laplace Transform with integral:

$$\int_0^{\infty } \frac{\text{erf}\left(\sqrt{t}\right)^3 \exp (-(s+1) t)}{\sqrt{t}} \, dt$$ Integrating with parts and substituting Sqrt[t]=x

$${\frac {\sqrt {\pi}}{\sqrt {s+1}}}-\int_{0}^{\infty }\!6\,{\frac { {\rm erf} \left(\sqrt {s+1}x\right) \left( {\rm erf} \left(x\right) \right) ^{2}{{\rm e}^{-{x}^{2}}}}{\sqrt {s+1}}}\,{\rm d}x $$

Consider the parametric integral : $$ I(a,b) = \int_{0}^{\infty}\!\!\!\text{erf}(ax)\,\text{erf}(bx)\,\text{erf}(bx)\,e^{-x^2}dx \tag{1}$$ and notice that:

$$\frac{\partial }{\partial a}\frac{\partial I(a,b)}{\partial b}=\int_0^{\infty } \frac{8 e^{-x^2-a^2 x^2-b^2 x^2} x^2 \text{erf}(b x)}{\pi } \, dx=\frac{4 b}{\pi ^{3/2} \left(a^2+b^2+1\right) \left(a^2+2 b^2+1\right)}+\frac{4 \tan ^{-1}\left(\frac{b}{\sqrt{a^2+b^2+1}}\right)}{\pi ^{3/2} \left(a^2+b^2+1\right)^{3/2}}$$ integrating over $b\in \{0,1\}$ and $a\in \{0,\sqrt{s+1}\}$

 Integrate[(4 b)/((1 + a^2 + b^2) (1 + a^2 + 2 b^2) \[Pi]^(3/2)) + (
 4 ArcTan[b/Sqrt[1 + a^2 + b^2]])/((1 + a^2 + b^2)^(3/2) \[Pi]^(
 /2)), {b, 0, 1}, Assumptions -> a > 0]
 (*(4 ArcTan[1/Sqrt[2 + a^2]])/((1 + a^2) Sqrt[2 + a^2] \[Pi]^(3/2))*)

 Integrate[(4 ArcTan[1/Sqrt[2 + a^2]])/((1 + a^2) Sqrt[2 + a^2] \[Pi]^(3/2)), {a,
 0, Sqrt[s + 1]}]
 (*----*)

$$\int_0^{\sqrt{s+1}} \frac{4 \tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2} \pi ^{3/2}} \, da$$ Connect everything together:

$$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)=\frac{4 \sqrt{\frac{1}{s+1}}}{s}-96\frac{\int_0^{\sqrt{s+1}} \frac{\tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2}} \, da}{\pi ^2 s \sqrt{s+1}}$$

.EDIT 15.02.2017.

Finding Integral:

$\int_0^{\sqrt{s+1}} \frac{\tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2}} \, da=\frac{1}{4} \pi \tan ^{-1}\left(\sqrt{s+1}\right)-\int_0^1 \frac{\tan ^{-1}\left(\frac{\sqrt{s+1}}{\sqrt{a^2+2}}\right)}{\sqrt{a^2+2} \left(a^2+1\right)} \, da$

$\int_0^1 \frac{\tan ^{-1}\left(\frac{\sqrt{s+1}}{\sqrt{a^2+2}}\right)}{\sqrt{a^2+2} \left(a^2+1\right)} \, da = \frac{1}{6} \left(3 \sqrt{s+1} X(s)+\pi \tan ^{-1}\left(\frac{\sqrt{s+1}}{\sqrt{3}}\right)\right) $

where X[s]is:

$X(s)=\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}+\sqrt{s+1} \sin (x)\right)} \, dx+\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}-\sqrt{s+1} \sin (x)\right)} \, dx$

Check numerics:

NIntegrate[ArcTan[Sqrt[s + 1]/Sqrt[x^2 + 2]]/(Sqrt[x^2 + 2]*(x^2 + 1)) /. 
s -> 1, {x, 0, 1}]
(* 0.396282 *)

X[s_?NumericQ] := 
NIntegrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
I Sqrt[2] + Sqrt[s + 1]* Sin[x]), {x, ArcSin[Sqrt[3]*Sqrt[2]/3], 
Pi/2}] + 
NIntegrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
I Sqrt[2] - Sqrt[s + 1]* Sin[x]), {x, ArcSin[Sqrt[3]*Sqrt[2]/3], 
Pi/2}]
1/6 (\[Pi] ArcTan[Sqrt[1 + s]/Sqrt[3]] + 3 Sqrt[1 + s] X[s]) /. 
s -> 1 // N
(* 0.396282 *)
Ok works.

Finding a integral :

$\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}-\sqrt{s+1} \sin (x)\right)} \, dx=?$

MMA has problem to caluculate this integral with limits,but without limits is: where a is Sqrt[s+1]

 Integrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
 I Sqrt[2] -a*Sin[x]), x]
 (*the result is a very long*)

deleted 2 characters in body

Source Link

edited Feb 15, 2017 at 17:03

Mariusz Iwaniuk

16.4k
1
27
46

Answer from comments.

Let $F(0)=\text{erf}\left(\sqrt{t}\right)^4$.Then $F'(t)=\frac{4 e^{-t} \text{erf}\left(\sqrt{t}\right)^3}{\sqrt{\pi } \sqrt{t}}$ and $F(0)=0$. $$\mathcal{L}_t\left[F'(t)\right](s)=s \left(\mathcal{L}_t[F(t)](s)\right)-F(0)$$ $$\mathcal{L}_t\left[\frac{4 e^{-t} \text{erf}\left(\sqrt{t}\right)^3}{\sqrt{\pi } \sqrt{t}}\right](s)=s \left(\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)\right)-0$$

$$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)=\frac{4 \left(\mathcal{L}_t\left[\frac{\text{erf}\left(\sqrt{t}\right)^3}{\sqrt{t}}\right](1+s)\right)}{\sqrt{\pi } s}$$ Solve Laplace Transform with integral:

$$\int_0^{\infty } \frac{\text{erf}\left(\sqrt{t}\right)^3 \exp (-(s+1) t)}{\sqrt{t}} \, dt$$ Integrating with parts and substituting Sqrt[t]=x

$${\frac {\sqrt {\pi}}{\sqrt {s+1}}}-\int_{0}^{\infty }\!6\,{\frac { {\rm erf} \left(\sqrt {s+1}x\right) \left( {\rm erf} \left(x\right) \right) ^{2}{{\rm e}^{-{x}^{2}}}}{\sqrt {s+1}}}\,{\rm d}x $$

Consider the parametric integral : $$ I(a,b) = \int_{0}^{\infty}\!\!\!\text{erf}(ax)\,\text{erf}(bx)\,\text{erf}(bx)\,e^{-x^2}dx \tag{1}$$ and notice that:

$$\frac{\partial }{\partial a}\frac{\partial I(a,b)}{\partial b}=\int_0^{\infty } \frac{8 e^{-x^2-a^2 x^2-b^2 x^2} x^2 \text{erf}(b x)}{\pi } \, dx=\frac{4 b}{\pi ^{3/2} \left(a^2+b^2+1\right) \left(a^2+2 b^2+1\right)}+\frac{4 \tan ^{-1}\left(\frac{b}{\sqrt{a^2+b^2+1}}\right)}{\pi ^{3/2} \left(a^2+b^2+1\right)^{3/2}}$$ integrating over $b\in \{0,1\}$ and $a\in \{0,\sqrt{s+1}\}$

 Integrate[(4 b)/((1 + a^2 + b^2) (1 + a^2 + 2 b^2) \[Pi]^(3/2)) + (
 4 ArcTan[b/Sqrt[1 + a^2 + b^2]])/((1 + a^2 + b^2)^(3/2) \[Pi]^(
 /2)), {b, 0, 1}, Assumptions -> a > 0]
 (*(4 ArcTan[1/Sqrt[2 + a^2]])/((1 + a^2) Sqrt[2 + a^2] \[Pi]^(3/2))*)

 Integrate[(4 ArcTan[1/Sqrt[2 + a^2]])/((1 + a^2) Sqrt[2 + a^2] \[Pi]^(3/2)), {a,
 0, Sqrt[s + 1]}]
 (*----*)

$$\int_0^{\sqrt{s+1}} \frac{4 \tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2} \pi ^{3/2}} \, da$$ Connect everything together:

$$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)=\frac{4 \sqrt{\frac{1}{s+1}}}{s}-96\frac{\int_0^{\sqrt{s+1}} \frac{4 \tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2}} \, da}{\pi ^2 s \sqrt{s+1}}$$$$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)=\frac{4 \sqrt{\frac{1}{s+1}}}{s}-96\frac{\int_0^{\sqrt{s+1}} \frac{\tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2}} \, da}{\pi ^2 s \sqrt{s+1}}$$

.EDIT 15.02.2017.

$\int_{0}^{1}\!{\frac {1}{ \left( {y}^{2}+1 \right) \sqrt {{y}^{2}+2}} \arctan \left( {\frac {\sqrt {s+1}}{\sqrt {{y}^{2}+2}}} \right) } \,{\rm d}y = \frac{1}{6} \left(3 \sqrt{s+1} X(s)+\pi \tan ^{-1}\left(\frac{\sqrt{s+1}}{\sqrt{3}}\right)\right) $

where X[s]is:

$X(s)=\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}+\sqrt{s+1} \sin (x)\right)} \, dx+\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}-\sqrt{s+1} \sin (x)\right)} \, dx$

Check numerics:

NIntegrate[ArcTan[Sqrt[s + 1]/Sqrt[x^2 + 2]]/(Sqrt[x^2 + 2]*(x^2 + 1)) /. 
s -> 1, {x, 0, 1}]
(* 0.396282 *)

X[s_?NumericQ] := 
NIntegrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
I Sqrt[2] + Sqrt[s + 1]* Sin[x]), {x, ArcSin[Sqrt[3]*Sqrt[2]/3], 
Pi/2}] + 
NIntegrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
I Sqrt[2] - Sqrt[s + 1]* Sin[x]), {x, ArcSin[Sqrt[3]*Sqrt[2]/3], 
Pi/2}]
1/6 (\[Pi] ArcTan[Sqrt[1 + s]/Sqrt[3]] + 3 Sqrt[1 + s] X[s]) /. 
s -> 1 // N
(* 0.396282 *)
Ok works.

Finding a integral :

$\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}-\sqrt{s+1} \sin (x)\right)} \, dx=?$

MMA has problem to caluculate this integral with limits,but without limits is: where a is Sqrt[s+1]

 Integrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
 I Sqrt[2] -a*Sin[x]), x]
 (*the result is a very long*)

Answer from comments.

Let $F(0)=\text{erf}\left(\sqrt{t}\right)^4$.Then $F'(t)=\frac{4 e^{-t} \text{erf}\left(\sqrt{t}\right)^3}{\sqrt{\pi } \sqrt{t}}$ and $F(0)=0$. $$\mathcal{L}_t\left[F'(t)\right](s)=s \left(\mathcal{L}_t[F(t)](s)\right)-F(0)$$ $$\mathcal{L}_t\left[\frac{4 e^{-t} \text{erf}\left(\sqrt{t}\right)^3}{\sqrt{\pi } \sqrt{t}}\right](s)=s \left(\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)\right)-0$$

$$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)=\frac{4 \left(\mathcal{L}_t\left[\frac{\text{erf}\left(\sqrt{t}\right)^3}{\sqrt{t}}\right](1+s)\right)}{\sqrt{\pi } s}$$ Solve Laplace Transform with integral:

$$\int_0^{\infty } \frac{\text{erf}\left(\sqrt{t}\right)^3 \exp (-(s+1) t)}{\sqrt{t}} \, dt$$ Integrating with parts and substituting Sqrt[t]=x

$${\frac {\sqrt {\pi}}{\sqrt {s+1}}}-\int_{0}^{\infty }\!6\,{\frac { {\rm erf} \left(\sqrt {s+1}x\right) \left( {\rm erf} \left(x\right) \right) ^{2}{{\rm e}^{-{x}^{2}}}}{\sqrt {s+1}}}\,{\rm d}x $$

Consider the parametric integral : $$ I(a,b) = \int_{0}^{\infty}\!\!\!\text{erf}(ax)\,\text{erf}(bx)\,\text{erf}(bx)\,e^{-x^2}dx \tag{1}$$ and notice that:

$$\frac{\partial }{\partial a}\frac{\partial I(a,b)}{\partial b}=\int_0^{\infty } \frac{8 e^{-x^2-a^2 x^2-b^2 x^2} x^2 \text{erf}(b x)}{\pi } \, dx=\frac{4 b}{\pi ^{3/2} \left(a^2+b^2+1\right) \left(a^2+2 b^2+1\right)}+\frac{4 \tan ^{-1}\left(\frac{b}{\sqrt{a^2+b^2+1}}\right)}{\pi ^{3/2} \left(a^2+b^2+1\right)^{3/2}}$$ integrating over $b\in \{0,1\}$ and $a\in \{0,\sqrt{s+1}\}$

 Integrate[(4 b)/((1 + a^2 + b^2) (1 + a^2 + 2 b^2) \[Pi]^(3/2)) + (
 4 ArcTan[b/Sqrt[1 + a^2 + b^2]])/((1 + a^2 + b^2)^(3/2) \[Pi]^(
 /2)), {b, 0, 1}, Assumptions -> a > 0]
 (*(4 ArcTan[1/Sqrt[2 + a^2]])/((1 + a^2) Sqrt[2 + a^2] \[Pi]^(3/2))*)

 Integrate[(4 ArcTan[1/Sqrt[2 + a^2]])/((1 + a^2) Sqrt[2 + a^2] \[Pi]^(3/2)), {a,
 0, Sqrt[s + 1]}]
 (*----*)

$$\int_0^{\sqrt{s+1}} \frac{4 \tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2} \pi ^{3/2}} \, da$$ Connect everything together:

$$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)=\frac{4 \sqrt{\frac{1}{s+1}}}{s}-96\frac{\int_0^{\sqrt{s+1}} \frac{4 \tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2}} \, da}{\pi ^2 s \sqrt{s+1}}$$

.EDIT 15.02.2017.

$\int_{0}^{1}\!{\frac {1}{ \left( {y}^{2}+1 \right) \sqrt {{y}^{2}+2}} \arctan \left( {\frac {\sqrt {s+1}}{\sqrt {{y}^{2}+2}}} \right) } \,{\rm d}y = \frac{1}{6} \left(3 \sqrt{s+1} X(s)+\pi \tan ^{-1}\left(\frac{\sqrt{s+1}}{\sqrt{3}}\right)\right) $

where X[s]is:

$X(s)=\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}+\sqrt{s+1} \sin (x)\right)} \, dx+\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}-\sqrt{s+1} \sin (x)\right)} \, dx$

Check numerics:

NIntegrate[ArcTan[Sqrt[s + 1]/Sqrt[x^2 + 2]]/(Sqrt[x^2 + 2]*(x^2 + 1)) /. 
s -> 1, {x, 0, 1}]
(* 0.396282 *)

X[s_?NumericQ] := 
NIntegrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
I Sqrt[2] + Sqrt[s + 1]* Sin[x]), {x, ArcSin[Sqrt[3]*Sqrt[2]/3], 
Pi/2}] + 
NIntegrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
I Sqrt[2] - Sqrt[s + 1]* Sin[x]), {x, ArcSin[Sqrt[3]*Sqrt[2]/3], 
Pi/2}]
1/6 (\[Pi] ArcTan[Sqrt[1 + s]/Sqrt[3]] + 3 Sqrt[1 + s] X[s]) /. 
s -> 1 // N
(* 0.396282 *)
Ok works.

Finding a integral :

$\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}-\sqrt{s+1} \sin (x)\right)} \, dx=?$

MMA has problem to caluculate this integral with limits,but without limits is: where a is Sqrt[s+1]

 Integrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
 I Sqrt[2] -a*Sin[x]), x]
 (*the result is a very long*)

Answer from comments.

Let $F(0)=\text{erf}\left(\sqrt{t}\right)^4$.Then $F'(t)=\frac{4 e^{-t} \text{erf}\left(\sqrt{t}\right)^3}{\sqrt{\pi } \sqrt{t}}$ and $F(0)=0$. $$\mathcal{L}_t\left[F'(t)\right](s)=s \left(\mathcal{L}_t[F(t)](s)\right)-F(0)$$ $$\mathcal{L}_t\left[\frac{4 e^{-t} \text{erf}\left(\sqrt{t}\right)^3}{\sqrt{\pi } \sqrt{t}}\right](s)=s \left(\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)\right)-0$$

$$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)=\frac{4 \left(\mathcal{L}_t\left[\frac{\text{erf}\left(\sqrt{t}\right)^3}{\sqrt{t}}\right](1+s)\right)}{\sqrt{\pi } s}$$ Solve Laplace Transform with integral:

$$\int_0^{\infty } \frac{\text{erf}\left(\sqrt{t}\right)^3 \exp (-(s+1) t)}{\sqrt{t}} \, dt$$ Integrating with parts and substituting Sqrt[t]=x

$${\frac {\sqrt {\pi}}{\sqrt {s+1}}}-\int_{0}^{\infty }\!6\,{\frac { {\rm erf} \left(\sqrt {s+1}x\right) \left( {\rm erf} \left(x\right) \right) ^{2}{{\rm e}^{-{x}^{2}}}}{\sqrt {s+1}}}\,{\rm d}x $$

Consider the parametric integral : $$ I(a,b) = \int_{0}^{\infty}\!\!\!\text{erf}(ax)\,\text{erf}(bx)\,\text{erf}(bx)\,e^{-x^2}dx \tag{1}$$ and notice that:

$$\frac{\partial }{\partial a}\frac{\partial I(a,b)}{\partial b}=\int_0^{\infty } \frac{8 e^{-x^2-a^2 x^2-b^2 x^2} x^2 \text{erf}(b x)}{\pi } \, dx=\frac{4 b}{\pi ^{3/2} \left(a^2+b^2+1\right) \left(a^2+2 b^2+1\right)}+\frac{4 \tan ^{-1}\left(\frac{b}{\sqrt{a^2+b^2+1}}\right)}{\pi ^{3/2} \left(a^2+b^2+1\right)^{3/2}}$$ integrating over $b\in \{0,1\}$ and $a\in \{0,\sqrt{s+1}\}$

 Integrate[(4 b)/((1 + a^2 + b^2) (1 + a^2 + 2 b^2) \[Pi]^(3/2)) + (
 4 ArcTan[b/Sqrt[1 + a^2 + b^2]])/((1 + a^2 + b^2)^(3/2) \[Pi]^(
 /2)), {b, 0, 1}, Assumptions -> a > 0]
 (*(4 ArcTan[1/Sqrt[2 + a^2]])/((1 + a^2) Sqrt[2 + a^2] \[Pi]^(3/2))*)

 Integrate[(4 ArcTan[1/Sqrt[2 + a^2]])/((1 + a^2) Sqrt[2 + a^2] \[Pi]^(3/2)), {a,
 0, Sqrt[s + 1]}]
 (*----*)

$$\int_0^{\sqrt{s+1}} \frac{4 \tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2} \pi ^{3/2}} \, da$$ Connect everything together:

$$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](s)=\frac{4 \sqrt{\frac{1}{s+1}}}{s}-96\frac{\int_0^{\sqrt{s+1}} \frac{\tan ^{-1}\left(\frac{1}{\sqrt{2+a^2}}\right)}{\left(1+a^2\right) \sqrt{2+a^2}} \, da}{\pi ^2 s \sqrt{s+1}}$$

.EDIT 15.02.2017.

$\int_{0}^{1}\!{\frac {1}{ \left( {y}^{2}+1 \right) \sqrt {{y}^{2}+2}} \arctan \left( {\frac {\sqrt {s+1}}{\sqrt {{y}^{2}+2}}} \right) } \,{\rm d}y = \frac{1}{6} \left(3 \sqrt{s+1} X(s)+\pi \tan ^{-1}\left(\frac{\sqrt{s+1}}{\sqrt{3}}\right)\right) $

where X[s]is:

$X(s)=\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}+\sqrt{s+1} \sin (x)\right)} \, dx+\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}-\sqrt{s+1} \sin (x)\right)} \, dx$

Check numerics:

NIntegrate[ArcTan[Sqrt[s + 1]/Sqrt[x^2 + 2]]/(Sqrt[x^2 + 2]*(x^2 + 1)) /. 
s -> 1, {x, 0, 1}]
(* 0.396282 *)

X[s_?NumericQ] := 
NIntegrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
I Sqrt[2] + Sqrt[s + 1]* Sin[x]), {x, ArcSin[Sqrt[3]*Sqrt[2]/3], 
Pi/2}] + 
NIntegrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
I Sqrt[2] - Sqrt[s + 1]* Sin[x]), {x, ArcSin[Sqrt[3]*Sqrt[2]/3], 
Pi/2}]
1/6 (\[Pi] ArcTan[Sqrt[1 + s]/Sqrt[3]] + 3 Sqrt[1 + s] X[s]) /. 
s -> 1 // N
(* 0.396282 *)
Ok works.

Finding a integral :

$\int_{\sin ^{-1}\left(\frac{\sqrt{3} \sqrt{2}}{3}\right)}^{\frac{\pi }{2}} \frac{i \left(\tan ^{-1}\left(\frac{2 \cos (x)}{\sin ^2(x)}\right) \cos (x)\right)}{2 \left(i \sqrt{2}-\sqrt{s+1} \sin (x)\right)} \, dx=?$

MMA has problem to caluculate this integral with limits,but without limits is: where a is Sqrt[s+1]

 Integrate[(I/2)*(ArcTan[2 *Cos[x]/Sin[x]^2] Cos[x])/(
 I Sqrt[2] -a*Sin[x]), x]
 (*the result is a very long*)

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edited Feb 15, 2017 at 14:20

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edited Feb 15, 2017 at 14:12

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edited Feb 7, 2017 at 14:50

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edited Feb 6, 2017 at 18:39

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edited Feb 6, 2017 at 11:13

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edited Feb 6, 2017 at 11:02

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created Feb 6, 2017 at 10:44

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