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I am trying to evaluate this integral?

$$\int_0^1\int_0^1 \frac{\mathrm dt\mathrm ds}{(c^2 + 2se + as^2 - 2tf - 2tsd + bt^2)^2}$$

I tried to do using the definite integral, but these two didn't converge:

Integrate[
 1/(c^2 + 2*s*e + a*s^2 - 2*t*f - 2*s*t*d + b*t^2)^2, {s, 0, 1}, {t, 
  0, 1}, Assumptions -> Element[a, Reals], 
 Assumptions -> Element[e, Reals], Assumptions -> Element[d, Reals], 
 Assumptions -> Element[b, Reals], Assumptions -> Element[f, Reals]]

and

Integrate[Integrate[
           1/(c^2 + 2*s*e + a*s^2 - 2*t*f - 2*s*t*d + b*t^2)^2, {s, 0, 1}, 
              Assumptions -> Element[a | e | d | f | b, Reals]], {t, 0, 1},   
              Assumptions -> Element[a | e | d | f | b, Reals]]

But if I do:

Integrate[
 Integrate[1/(c^2 + 2*s*e + a*s^2 - 2*t*f - 2*s*t*d + b*t^2)^2, s, 
  Assumptions -> Element[a | e | d | f | b, Reals]], t, 
 Assumptions -> Element[a | e | d | f | b, Reals]]

surprisingly, it works very fast and this is the result:

(((-b (e + a s) + d (f + d s)) ArcTan[(f + d s - b t)/
   Sqrt[-(f + d s)^2 + b (c^2 + 2 e s + a s^2)]])/Sqrt[-(f + d s)^2 + 
  b (c^2 + 2 e s + a s^2)] + ((d e - a f + a b t - d^2 t) ArcTan[(
   e + a s - d t)/
   Sqrt[-(e - d t)^2 + a (c^2 - 2 f t + b t^2)]])/Sqrt[-(e - d t)^2 + 
  a (c^2 - 2 f t + b t^2)])/(2 (c^2 d^2 + e (b e - 2 d f) + 
   a (-b c^2 + f^2)))

Now, how can I assign the indefinite integral expression into a function and evaluate it at 0 and 1 using Mathematica?

F[s_, t_] := 
 Integrate[
   Integrate[1/(c^2 + 2*s*e + a*s^2 - 2*t*f - 2*s*t*d + b*t^2)^2, s, 
    Assumptions -> Element[a | e | d | f | b, Reals]], t, 
   Assumptions -> Element[a | e | d | f | b, Reals]] // FullSimplify

Doesn't actually evaluate the integral upon declaration. How can I feed Integrate into a function.

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  • 3
    $\begingroup$ Multiple Assumptions options are not going to work. I'm not really giving high hopes on improving calculation time, but I'd suggest writing this as Integrate[1/(c^2 + 2*s*e + a*s^2 - 2*t*f - 2*s*t*d + b*t^2)^2, {s, 0, 1}, {t, 0, 1}, Assumptions -> Element[a | b | d | e | f, Reals]]. (You could also group them as Element[a, Reals] && Element[b, Reals] && ....) Did you possibly forget to include c in the assumptions? $\endgroup$ – kirma Sep 18 '18 at 15:18
  • $\begingroup$ @kirma doesn't finish without the assumptions either. $\endgroup$ – 0x90 Sep 18 '18 at 15:22
  • 1
    $\begingroup$ What does "doesn't finish" mean? Not all integrals, especially complicated symbolic ones, can be calculated in an instant (if at all). $\endgroup$ – user6014 Sep 18 '18 at 15:23
  • $\begingroup$ I would guess complete assumptions could actually speed up the operation. Your variant of listing multiple Assumptions options is going to cause Integrate pick only one of them. $\endgroup$ – kirma Sep 18 '18 at 15:27
  • 1
    $\begingroup$ Try your integral in Rubi. Perhaps that can solve it. $\endgroup$ – user6014 Sep 18 '18 at 15:28
5
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I use Rule-based Integration aka: Rubi from here:

<<Rubi` (*Load package*)

s1 = Int[1/(c^2 + 2*s*e + a*s^2 - 2*t*f - 2*s*t*d + b*t^2)^2, t]
s2 = Limit[s1, t -> 1] - Limit[s1, t -> 0] // Simplify
(*I assumed the original function is continuous*)

s3= Int[s2, s]

s4 = Limit[s3, s -> 1, Assumptions -> {e > 0, c > 0, a > 0, f > 0, d > 0, b > 0}] - Limit[s3, s -> 0, 
Assumptions -> {e > 0, c > 0, a > 0, f > 0, d > 0, b > 0}]
(*I assumed the original function is continuous. *)

(*Warning!!! See restrictions about parameters ? *)

s5 = s4[[1]] // Simplify
(* Solution *)

Formula for:

$e > 0, c > 0, a > 0, f > 0, d > 0, b > 0$

and

$-1<\frac{(b e-d f) \Im\left(\frac{1}{\sqrt{a b c^2-c^2 d^2-b e^2+2 d e f-a f^2}}\right)}{\sqrt{b}}<1\land b c^2>f^2\land (d+f)^2<b \left(a+c^2+2 e\right)$

SOL[a_, c_, e_, f_, d_, b_] := (1/(2 (a b c^2 - c^2 d^2 - b e^2 + 2 d e f - 
a f^2)))(((d e - a f) ArcTan[e/Sqrt[a c^2 - e^2]])/Sqrt[
a c^2 - e^2] + ((-d e + a f) ArcTan[(a + e)/Sqrt[a c^2 - e^2]])/
Sqrt[a c^2 - 
e^2] + ((d (-d + e) + a (b - f)) ArcTan[(d - e)/
Sqrt[-(d - e)^2 + a (b + c^2 - 2 f)]])/
Sqrt[-(d - e)^2 + 
a (b + c^2 - 2 f)] + ((d (d - e) + a (-b + f)) ArcTan[(-a + d - e)/
Sqrt[-(d - e)^2 + a (b + c^2 - 2 f)]])/
Sqrt[-(d - e)^2 + 
a (b + c^2 - 2 f)] + ((-b e + d f) ArcTan[(b - f)/Sqrt[
b c^2 - f^2]])/Sqrt[
b c^2 - f^2] + ((-b e + d f) ArcTan[f/Sqrt[b c^2 - f^2]])/Sqrt[
b c^2 - f^2] + ((a b - d^2 + b e - 
d f) (ArcTan[(b - d - f)/Sqrt[b (a + c^2 + 2 e) - (d + f)^2]] + 
ArcTan[(d + f)/Sqrt[b (a + c^2 + 2 e) - (d + f)^2]]))/Sqrt[
a b + b (c^2 + 2 e) - (d + f)^2]);

At first we must check restrictions:

-1 < ((b e - d f) Im[1/Sqrt[a b c^2 - c^2 d^2 - b e^2 + 2 d e f - a f^2]])/Sqrt[b] < 1 && 
 b c^2 > f^2 && (d + f)^2 < b (a + c^2 + 2 e) /. {a -> 1, c -> 2, e -> 1, f -> 2, d -> 1, b -> 2}
 (*True *)(*OK*) 

If false then forumla gives complex number (Wrong answer!).

then:

 F[a_, c_, e_, f_, d_, b_] := NIntegrate[1/(c^2 + 2*s*e + a*s^2 - 2*t*f - 2*s*t*d + b*t^2)^2, {t, 0, 1}, {s, 0, 1}]
 F[1, 2, 1, 2, 1, 2]
 (* 0.105123 *)

 SOL[1, 2, 1, 2, 1, 2] // N
 (* 0.105123 *)

Works as should it.

For some values use Limit,because formula gives Power::infy: Infinite expression 1/0 encountered

At first check restrictions about parameters:

 1 < ((b e - d f) Im[1/Sqrt[a b c^2 - c^2 d^2 - b e^2 + 2 d e f - a f^2]])/Sqrt[b] <
 1 && b c^2 > f^2 && (d + f)^2 < b (a + c^2 + 2 e) /. {a -> 1, e -> 1, f -> 
 1, d -> 1, b -> 2}

 (*-1 < Im[1/Sqrt[-1 + c^2]]/Sqrt[2] < 1 && 2 c^2 > 1 && 4 < 2 (3 + c^2)*)

 Limit[Im[1/Sqrt[-1 + c^2]]/Sqrt[2], c -> 1, Direction -> -1]
 (* 0 *)
 -1 < 0 < 1 && 2 c^2 > 1 && 4 < 2 (3 + c^2) /. c -> 1
 (*True*)

 F[1, 1, 1, 1, 1, 2]
 (* 0.803374 *)
 SOL[1, 1, 1, 1, 1, 2](* errors *)

 Limit[SOL[1, c, 1, 1, 1, 2], c -> 1] // N
 (* 0.803374 *)

Bonus Extra:

With a small changes:

s4 = Limit[s3, s -> 1, Assumptions -> {e ∈ Reals, c ∈ Reals, a ∈ Reals, f ∈ Reals, d ∈ Reals, b ∈ Reals}]-
Limit[s3, s -> 0, Assumptions -> {e ∈ Reals, c ∈ Reals, a ∈ Reals, f ∈ Reals, 
d ∈ Reals,b ∈ Reals}]//Simplify
(* solution *)

SOL2[a_, c_, e_, f_, d_, b_] := (1/(
2 (a b c^2 - c^2 d^2 - b e^2 + 2 d e f - 
 a f^2)))(((d e - a f) ArcTan[e/Sqrt[a c^2 - e^2]])/Sqrt[
a c^2 - e^2] + ((-d e + a f) ArcTan[(a + e)/Sqrt[a c^2 - e^2]])/
Sqrt[a c^2 - 
 e^2] + ((d (-d + e) + a (b - f)) ArcTan[(d - e)/
  Sqrt[-(d - e)^2 + a (b + c^2 - 2 f)]])/
Sqrt[-(d - e)^2 + 
 a (b + c^2 - 2 f)] + ((d (d - e) + a (-b + f)) ArcTan[(-a + d - 
   e)/Sqrt[-(d - e)^2 + a (b + c^2 - 2 f)]])/
Sqrt[-(d - e)^2 + 
 a (b + c^2 - 2 f)] + ((-b e + 
   d f) (ArcTan[(b - f)/Sqrt[b c^2 - f^2]] + 
   ArcTan[f/Sqrt[b c^2 - f^2]]))/Sqrt[
 b c^2 - f^2] + ((a b - d^2 + b e - 
   d f) (ArcTan[(b - d - f)/Sqrt[
    a b + b (c^2 + 2 e) - (d + f)^2]] + 
   ArcTan[(d + f)/Sqrt[a b + b (c^2 + 2 e) - (d + f)^2]]))/Sqrt[
 a b + b (c^2 + 2 e) - (d + f)^2]);

$$\int _0^1\int _0^1\frac{1}{\left(c^2+2 s e+a s^2-2 t f-2 s t d+b t^2\right)^2}dtds=\\\frac{\frac{(d e-a f) \tan ^{-1}\left(\frac{e}{\sqrt{a c^2-e^2}}\right)}{\sqrt{a c^2-e^2}}+\frac{(-d e+a f) \tan ^{-1}\left(\frac{a+e}{\sqrt{a c^2-e^2}}\right)}{\sqrt{a c^2-e^2}}+\frac{(d (-d+e)+a (b-f)) \tan ^{-1}\left(\frac{d-e}{\sqrt{-(d-e)^2+a \left(b+c^2-2 f\right)}}\right)}{\sqrt{-(d-e)^2+a \left(b+c^2-2 f\right)}}+\frac{(d (d-e)+a (-b+f)) \tan ^{-1}\left(\frac{-a+d-e}{\sqrt{-(d-e)^2+a \left(b+c^2-2 f\right)}}\right)}{\sqrt{-(d-e)^2+a \left(b+c^2-2 f\right)}}+\frac{(-b e+d f) \left(\tan ^{-1}\left(\frac{b-f}{\sqrt{b c^2-f^2}}\right)+\tan ^{-1}\left(\frac{f}{\sqrt{b c^2-f^2}}\right)\right)}{\sqrt{b c^2-f^2}}+\frac{\left(a b-d^2+b e-d f\right) \left(\tan ^{-1}\left(\frac{b-d-f}{\sqrt{a b+b \left(c^2+2 e\right)-(d+f)^2}}\right)+\tan ^{-1}\left(\frac{d+f}{\sqrt{a b+b \left(c^2+2 e\right)-(d+f)^2}}\right)\right)}{\sqrt{a b+b \left(c^2+2 e\right)-(d+f)^2}}}{2 \left(a b c^2-c^2 d^2-b e^2+2 d e f-a f^2\right)}$$

Formula for:

$\{e\in \mathbb{R},c\in \mathbb{R},a\in \mathbb{R},f\in \mathbb{R},d\in \mathbb{R},b\in \mathbb{R}\}$

with restrictions:

$b c^2-f^2>0\land (d+f)^2<b \left(a+c^2+2 e\right)$

Reduce[b c^2 > f^2 && (d + f)^2 < b (a + c^2 + 2 e)]

$$(d|e|f)\in \mathbb{R}\land \left(\left(c<0\land b>\frac{f^2}{c^2}\land a>\frac{-b c^2-2 b e+d^2+2 d f+f^2}{b}\right)\lor \left(c>0\land b>\frac{f^2}{c^2}\land a>\frac{-b c^2-2 b e+d^2+2 d f+f^2}{b}\right)\right)$$

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