# What's my mistake in finding $\int_0^\infty dx e^{-ax^2} \sin(b/x^2)$?

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.

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

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

Therefore

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

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