How to evaluate $\int_{-\infty}^{\infty} e^{-k^{2} t^{2}-\frac{z^2}{4t^{2}}} dt$

Question

I try to solve the integral by substitute $u = t - \frac{z}{2kt}$ then I could get:

$$\int_{-\infty}^{\infty} e^{-k^{2} t^{2}-\frac{z^2}{4t^{2}}} dt = e^{-kz} \int_{-\infty}^{\infty} e^{-u^2} dt$$

However, I am not sure about how to solve the remaining integral because I can't express $u$ in terms of $t$ explicitly.

I would be appreciated if anyone have a comment about it.

Answer 1

Rewrite the integral$$I=\int_{-\infty}^{\infty} e^{-k^{2} t^{2}-\frac{z^2}{4t^{2}}} dt= e^{-{kz}}\int_{-\infty}^{\infty} e^{-k^{2} (t-\frac{z}{2kt})^2} dt $$ and then apply the Glasser's master theorem $\int_{-\infty}^{\infty} f(x-a/x) dx=\int_{-\infty}^{\infty} f(x) dx$ to arrive at $$I= e^{-{kz}}\int_{-\infty}^{\infty} e^{-k^{2} t^2} dt =\frac{\sqrt{\pi}}{|k|}e^{-{kz}} $$

Answer 2

Without using Glasser's Master Theorem.

The core is to compute $$F(b)=\int_0^\infty e^{-x^2-\frac{b^2}{x^2}}~dx=\int_0^\infty e^{-\left(x-\frac{b}{x}\right)^2-2b}dx$$

Without loss of generality, let $b\ge0$, we get

$$e^{2b}F(b)=\int_0^\infty e^{-\left(x-\frac{b}{x}\right)^2}dx\tag{1}$$ Take derivative $$F'(b)=\int_0^\infty e^{x^2-\frac{b^2}{x^2}}\cdot\frac{-2b}{x^2}~dx=2\int_0^\infty e^{x^2-\frac{b^2}{x^2}}~d\left(\frac{b}{x}\right)=2\int_0^\infty e^{-\left(x-\frac{b}x\right)^2-2b}~~d\left(\frac{b}{x}\right)$$ we get

$$\frac{1}2e^{2b}F'(b)=\int_0^\infty e^{-\left(x-\frac{b}x\right)^2}~d\left(\frac{b}{x}\right)\tag{2}$$

Let $(1)-(2)$

$$e^{2b}F-\frac{1}2e^{2b}F'=\int_0^\infty e^{-\left(x-\frac{b}x\right)^2}d\left(x-\frac{b}{x}\right)$$

Let $u=x-\frac{b}{x}$

$$e^{2b}F-\frac{1}2e^{2b}F'=\int_{-\infty}^\infty e^{-u^2}du=\sqrt\pi$$

So we get a first order differential equation

$$F-\frac{1}2F'=\sqrt\pi e^{-2b},~~~~F(0)=\frac{\sqrt\pi}{2}$$

So we get

$$\boxed{\int_0^\infty e^{-x^2-\frac{b^2}{x^2}}~dx=F(b)=\frac{\sqrt\pi}{2}e^{-2b},~~~b\ge0}$$

Finally,

$$I=\int_{-\infty}^{\infty} e^{-k^{2} t^{2}-\frac{z^2}{4t^{2}}} dt=2\int_0^{\infty} e^{-k^{2} t^{2}-\frac{z^2}{4t^{2}}} dt$$

Let $x=|k|t$

$$I=\frac{2}{|k|}\int_0^{\infty} e^{-x^{2}-\frac{z^2k^2/4}{x^{2}}} dx$$

Plut in $b=\frac{|zk|}2$

$$\boxed{I=\frac{\sqrt\pi}{|k|}e^{-|zk|}}$$