1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ After the interchange, all the integrals diverge at $0$, aren't they? $\endgroup$ – user58697 Jul 29 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 at 20:09
1
$\begingroup$

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

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.