$$\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{\mathrm dx\mathrm dy}{xy}\sinh(\pi x)\sinh(\pi y)e^{\pi i(2x^2+y^2)}\sin(\pi x y)$$

Here both integration variables $x$ and $y$ run from $-\infty$ to $\infty$. I have tried the following code, but it is answering zero, which is not the correct answer. An answer is a non-zero number something 0.8, might be a complex number.

my code

Integrate[(Sinh[Pi*x]*Sinh[Pi*y]*Sin[Pi*x*y]*E^(Pi*I*(2*x^2 + y^2)))/
   (x*y), {x, -Infinity, Infinity}, {y, -Infinity, Infinity}]
  • 1
    $\begingroup$ please post plain text Mathematica code as well, so one can copy and paste it. A little hard to copy code from an image. $\endgroup$
    – Nasser
    Apr 14, 2020 at 12:03
  • $\begingroup$ Think of NIntegrate[(Sinh[Pi * x]*Sinh[Pi * y]*Sin[Pi * x * y] * E^(Pi * I*(2 * x^2 + y^2)))/(x * y) /. y -> 1, {x, -100, 200}] which results in $ -1.07749\times 10^{265}-2.68302\times 10^{262} i$. This suggests the divergence of the improper double integral under consideration. $\endgroup$
    – user64494
    Apr 14, 2020 at 13:24
  • $\begingroup$ Executing Integrate[(Sinh[Pi * x]*Sinh[Pi * y] * Sin[Pi x * y]*E^(Pi * I (2 * x^2 + y^2)))/ (x * y), {x, -Infinity, Infinity}, {y, -Infinity, Infinity}] in version 12.0, one obtains an error communication "integral of (E^(I [Pi] (2 x^2+y^2)) Sin[[Pi] x y] Sinh[[Pi] x] Sinh[[Pi] y])/(x y) does not converge on {-[Infinity],[Infinity]}" and the returned input. $\endgroup$
    – user64494
    Apr 14, 2020 at 14:52
  • $\begingroup$ Both en.wikipedia.org/wiki/Cauchy_principal_value as well as encyclopediaofmath.org/index.php/Improper_integral say only about multidimensional integrals where the integrand has an isolated singularity at a point. In view of it the integral from the question cannot be treated as its Cauchy principal value. Therefore, 0 because of the oddity of the integrand in x is a wrong answer. $\endgroup$
    – user64494
    Apr 14, 2020 at 15:44

1 Answer 1


The solution zero is true, since the integrand is anti-symmetric in x and y.

integrand[x_, y_] = 
 E^(Pi*I*(2*x^2 + y^2)))/(x*y);

integrand[x, y] + integrand[-x, y]

yields zero.

  • 1
    $\begingroup$ Thanks, need no explanation from you. If integrand is anti-symmetric, every integral from -a to a or - inf to inf yields zero. $\endgroup$
    – Akku14
    Apr 14, 2020 at 13:56
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    $\begingroup$ Convergence at infinity is not obvious and should be validated in some way. $\endgroup$ Apr 14, 2020 at 14:19
  • 1
    $\begingroup$ No, it's not a Cauchy principal value singular integral. The issue is that it is unclear whether the oscillation is sufficiently fast to handle the growth at infinity. A way to check convergence is to get a first-order approximation of the result from one period of oscillation, as a function of either variable. Then check whether summing such terms will converge. $\endgroup$ Apr 14, 2020 at 16:31
  • 1
    $\begingroup$ As an addendum to my comments, I will mention that I am familiar with the convergence testing code in Integrate. More than anyone else, in fact. So take this as tacit admission that it is by no means without flaws. This is very much an issue when it comes to assessing oscillatory integrands at infinity. $\endgroup$ Apr 14, 2020 at 16:34
  • 1
    $\begingroup$ By it I refer to the integral in question. As for the rest, I pass (We are still working, albeit remotely, and I do have things I need to do. And maybe so do you.) $\endgroup$ Apr 15, 2020 at 15:02

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