I need to find the integration of the following expression


But I am getting some syntax error.

If $\delta=1/2$ and $\delta=2/3$, how to solve these symbolically or numerically?


closed as off-topic by Daniel Lichtblau, Jens, RunnyKine, MarcoB, user9660 Mar 29 '16 at 4:19

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  • $\begingroup$ Welcome! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! $\endgroup$ – user9660 Mar 28 '16 at 17:01
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  • $\begingroup$ Obviously I have a different understanding, but I don't agree with the off-topic verdict. The problem is (almost) clearly stated, the missing information on k and t0 is easily completed, and the solution is IMHO a non-trivial exercise in Mathematica. It even led to me the discovery of a bug in Mathematica. $\endgroup$ – Dr. Wolfgang Hintze Mar 29 '16 at 17:24

We calculate the symbolic result of the integral for the two values of the parameter δ.

Considering, as I did before in a related problem (Strange result of parameter-dependent definite integral), the exponential form instead of the cosine I obtain after expanding the exponential function containig δ and integrating term by term the following result (where t = τ0)

H[δ_] = 
  1/(π k I )
    Sum[(-1)^n/n! Gamma[n δ + 1]/(-k t )^(n*δ), {n, 
     0, ∞}];

Interestingly Mathematica can calculate the sum explicity for the two values of δ requested:


$$-\frac{i e^{-\frac{1}{4 k t}} \left(-\sqrt{\pi } \sqrt{-k t} \text{erfi}\left(\frac{1}{2} \sqrt{\frac{1}{k t}}\right)+\sqrt{\pi } (-k) t \sqrt{\frac{1}{k t}}+2 k t e^{\frac{1}{4 k t}} \sqrt{-k t} \sqrt{\frac{1}{k t}}\right)}{2 \pi k^2 t \sqrt{\frac{1}{k t}} \sqrt{-k t}}$$

FullSimplify[%, {t > 0, k > 0}]

$$\text{h12}=\frac{\frac{2 i F\left(\frac{1}{2 \sqrt{k t}}\right)}{\sqrt{k t}}+\frac{\sqrt{\pi } e^{-\frac{1}{4 k t}}}{\sqrt{k t}}-2 i}{2 \pi k}$$

Here F() is the function DawsonF[].

The real part is

$$\text{h12r}=\frac{e^{-\frac{1}{4 k t}}}{2 \sqrt{\pi } k \sqrt{k t}};$$

And with the next parameter value we find

h23 = FullSimplify[H[2/3], {t > 0, k > 0}]

$$-\frac{i t \left(-9 (-k t)^{4/3} \, _2F_2\left(\frac{1}{2},1;\frac{1}{3},\frac{2}{3};-\frac{4}{27 k^2 t^2}\right)+6 \Gamma \left(\frac{2}{3}\right) (-k t)^{2/3} \, _1F_1\left(\frac{5}{6};\frac{2}{3};-\frac{4}{27 k^2 t^2}\right)-2 \Gamma \left(\frac{1}{3}\right) \, _1F_1\left(\frac{7}{6};\frac{4}{3};-\frac{4}{27 k^2 t^2}\right)\right)}{9 \pi (-k t)^{7/3}}$$

The real part can be simplified to

h23r = (t/(π Sqrt[3]) ) ((
     Gamma[2/3] Hypergeometric1F1[5/6, 2/
       3, -(4/(27 k^2 t^2))])/(k t)^(5/3) + (
     Gamma[4/3] Hypergeometric1F1[7/6, 4/
       3, -(4/(27 k^2 t^2))])/(k t)^(7/3));

In Latex

$$\text{h23r}=\frac{t \left(\frac{\Gamma \left(\frac{2}{3}\right) \, _1F_1\left(\frac{5}{6};\frac{2}{3};-\frac{4}{27 k^2 t^2}\right)}{(k t)^{5/3}}+\frac{\Gamma \left(\frac{4}{3}\right) \, _1F_1\left(\frac{7}{6};\frac{4}{3};-\frac{4}{27 k^2 t^2}\right)}{(k t)^{7/3}}\right)}{\pi \sqrt{3}};$$

The graphs are

Plot[{h12r /. k -> 1/2, Re[h23 /. k -> 1/2]}, {t, 0, 5}, 
 PlotLabel -> 
  "Two integrals for k = 1/2\nblue curve -> δ = 1/2, yellow curve -> \
δ = 2/3", AxesLabel -> {"t", "H"}]

enter image description here


Honestly speaking, this kind of question does not have a definite answer. Whenever you are trying to do such integration, first thing you should check is whether your function is convergent or not. Considering your function, it will not converge for any parameter values.

So I have to give a generic answer. If you don't know the properties of your function, evaluate it on the parameter space and then plot it. You might get an idea of the parameter dependence of the integral.

For your example

f[d_, k_, t0_, u_] := t0/Pi Exp[-u^d Cos[d Pi/2]] Cos[u^d Sin[d Pi/2] - k t0 u]

Does not converge for d=1

NIntegrate[ f[1, 1, 1, u], {u, 0, Infinity}]


So choose smaller d and evaluate it

dat1 = Table[{k, t0,
  f[.1, k, t0, u], {u, 0, Infinity}]}
 ,{k, -0.5, .5, .1}, {t0, 0.,1., .1}];
dat2 = Table[{k, t0,
  f[.5, k, t0, u], {u, 0, Infinity}]}
 ,{k, -0.5, .5, .1}, {t0, 0.,1., .1}];

dat1 = Flatten[dat1, 1];
dat2 = Flatten[dat2, 1];

ListPlot3D[dat1, PlotLabel -> "d=0.1"]
ListPlot3D[dat2, PlotLabel -> "d=0.5"]

enter image description here

For advanced integration techniques you might be interested in this.

Now let say you want the functional relationship. So you can use Interpolation.

g = Interpolation[dat2]
Plot3D[g[x, y], {x, -0.5, 0.5}, {y, 0, 1}]

enter image description here

g[k,t0] is your approximate solution for d=0.5.


You have no assumptions made for tau and k.

f = t/Pi Exp[-u^d Cos[d Pi/2]] Cos[u^d Sin[d Pi/2] - k t u];
d = 1/2;

1.) solution:

int1 = Integrate[f, {u, 0, \[Infinity]}, Assumptions -> k*t != 0 && k*t \[Element] Reals]

enter image description here

2.) solution:

f1 = f // TrigToExp

enter image description here

int2 = Integrate[f1, {u, 0, \[Infinity]}, Assumptions -> k*t != 0 && k*t \[Element] Reals]
(* 0 *)

We may believe what is right! The plots point to 0.

{Plot3D[int1, {k, -5, -0.1}, {t, -5, -0.1}, PlotRange -> All, PlotPoints -> 100], 
 Plot3D[int1, {k, 0.1, 5}, {t, 0.1, 5}, PlotRange -> All, PlotPoints -> 100]}

enter image description here

{Plot3D[int1, {k, -5, -0.1}, {t, 0.1, 5}, PlotRange -> All, PlotPoints -> 100], 
 Plot3D[int1, {k, 0.1, 5}, {t, -5, -0.1}, PlotRange -> All, PlotPoints -> 100]}

enter image description here


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