I want to evaluate $$I=\int_0^\infty dx e^{-ax^2} \sin(b/x^2)$$ for $a,b>0$. A first simplification is to substitute $y=x/\sqrt{a}$ and define $c=ab>0$ to obtain $$I=\frac{1}{\sqrt{a}} \int_0^\infty e^{-x^2} \sin(c/x^2)$$ Now my idea was to use the Taylor series for the sine $$I=a^{-1/2} \int_0^\infty dx e^{-x^2} \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}\left(\frac{c}{x^2} \right)^{2k+1}$$ Now I interchange sum and integral although I have no justification $$ I=a^{-1/2} \sum_{k=0}^\infty \frac{(-1)^kc^{2k+1}}{(2k+1)!} \int_0^\infty dx e^{-x^2} \left(\frac{1}{x^2} \right)^{2k+1}$$ Substituting $t=x^2$ in the integral we obtain the gamma function $$I=a^{-1/2} \sum_{k=0}^\infty \frac{(-1)^kc^{2k+1}}{(2k+1)!} \frac{1}{2} \Gamma(-2k-1/2) $$ Using $\Gamma \left({\frac{1}{2}}-n\right)={(-4)^{n}n! \over (2n)!}{\sqrt {\pi }}$ (which can be shown using the reflection formula and the duplication formula for the Gamma function) with $n=2k+1$ I obtain $$I=a^{-1/2} \sqrt{\pi} \sum_{k=0}^\infty \frac{(-1)^k(-4c)^{2k+1}}{(4k+2)!} \frac{1}{2} $$ or $$I=-\frac{1}{2}\sqrt{\frac{\pi}{a}} \sum_{k=0}^\infty \frac{(-1)^k(4c)^{2k+1}}{(4k+2)!}=-\frac{1}{2}\sqrt{\frac{\pi}{a}} \sin(\sqrt{2c}) \sinh(\sqrt{2c})$$ where I used wolfram alpha for the last series.

The problem: The above result is wrong. It should be (wolfram alpha and Gradshteyn) $$I=\frac{1}{2}\sqrt{\frac{\pi}{a}} \sin(\sqrt{2c})\exp(-\sqrt{2c}) $$

The question: Can someone spot my mistake? Was it interchanging the limits? I would also be interested in your solutions to the integral $I$ using other approaches.

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    $\begingroup$ After the interchange, all the integrals diverge at $0$, aren't they? $\endgroup$ – user58697 Jul 29 '19 at 19:58
  • $\begingroup$ @user58697 Yes, you are right. Probably that is the mistake... Do you have an idea for an alternative approach to the original integral? $\endgroup$ – thomasfermi Jul 29 '19 at 20:09

The Glasser's master theorem is a useful tool for the solution. First, use Euler's formula to decompose the sine term into the sum of exponentials. Then it boils down to computing the integral of the form

$$ J(p) = \int_{0}^{\infty} \exp\left( -a x^2 - \frac{p}{x^2} \right) \, \mathrm{d}x. $$

Assume for a moment that $a, p > 0$. Then by completing the square, we get

$$ J(p) = \int_{0}^{\infty} \exp\left( -a \left( x - \frac{\smash{\sqrt{p/a}}}{x} \right)^2 - 2\sqrt{ap} \right) \, \mathrm{d}x. $$

Then by the Glasser's master theorem and the gaussian integral, this evaluates to

$$ J(p) = \int_{0}^{\infty} \exp\left( -a x^2 - 2\sqrt{ap} \right) \, \mathrm{d}x = \frac{1}{2}\sqrt{\frac{\pi}{a}} \exp(-2\sqrt{ap}). \tag{*}$$

Although $\text{(*)}$ is originally proved for $p > 0$, both sides of $\text{(*)}$ define holomorphic functions for $p$ in the right-half plane $\mathbb{H}_{\to} = \{z \in \mathbb{C} : \operatorname{Re}(z) > 0\}$ and are continuous on the closed right-half plane $\overline{\mathbb{H}_{\to}}$. So by the identity theorem and continuity, $\text{(*)}$ extends to all of $p \in \overline{\mathbb{H}_{\to}}$. In particular, plugging $p = \pm ib$ for $b > 0$, we get

$$ J(\pm ib) = \frac{1}{2}\sqrt{\frac{\pi}{a}} \exp(-2\sqrt{\pm i ab}) = \frac{1}{2}\sqrt{\frac{\pi}{a}} \exp(-\sqrt{2c}(1\pm i)). $$


$$ I = \frac{J(-ib) - J(ib)}{2i} = \frac{1}{2}\sqrt{\frac{\pi}{a}} \sin(\sqrt{2c}) \exp(-\sqrt{2c}). $$

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  • $\begingroup$ Thanks for your very nice solution. I had never heard of Glasser's theorem before! $\endgroup$ – thomasfermi Jul 30 '19 at 22:03

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