Simplifying integral:$\int_{0}^{\infty}{\exp\left(-\left(u^2+{ {\alpha^2}\over {16u^2t}}\right)\right)}~\mathrm{d}u$

Question

$$I(t)=\int_{0}^{\infty}{\exp\left(-\left(u^2+{ {\alpha^2}\over {16u^2t}}\right)\right)}~\mathrm{d}u $$

where $\alpha$ and $t$ are positive constant.

P.S.I would like to edit this problem, because I found my mistake about inner sign.

Thanks!

Answer 1

Hint. For any $f$ such that $\displaystyle \int_{-\infty}^{+\infty} f(u)\: \mathrm{d}x<\infty$, we have (see a proof here)

$$ \int_{-\infty}^{+\infty}f\left(u-\frac{s}{u}\right)\mathrm{d}u=\int_{-\infty}^{+\infty} f(u)\: \mathrm{d}u, \quad s>0. \tag1 $$

Apply it to $f(u)=e^{-u^2}$, you get

$$ \int_{-\infty}^{+\infty}e^{-(u-s/u)^2}\mathrm{d}u=\int_{-\infty}^{+\infty} e^{-u^2} \mathrm{d}u=\sqrt{\pi}, \quad s>0. \tag2 $$

or

$$ \int_{-\infty}^{+\infty}e^{-u^2-s^2/u^{2}}\mathrm{d}u=\sqrt{\pi}\:e^{-2s}\tag3 $$ then put $s=\dfrac{\alpha}{4\sqrt{t}}$ to obtain the value of your integral, noticing that $f$ is even here.

Answer 2

Using $$x^{2} - \frac{a^{2}}{x^{2}} = \left(x - \frac{i a}{x} \right)^{2} + 2 i a $$ then \begin{align} I &= \int_{0}^{\infty} e^{- \left(u^{2} - \frac{\alpha^{2}}{16 \, t \, u^{2}}\right)} \, du = e^{-2 i a} \, \int_{0}^{\infty} e^{- \left(u - \frac{i a}{u}\right)^{2}} \, du \end{align} where $4 \sqrt{t} \, a = \alpha$. Let $u = \sqrt{i a} x$ to obtain \begin{align} \tag{1} e^{2 i a} \, I = \sqrt{i a} \, \int_{0}^{\infty} e^{- i a \, \left(x - \frac{1}{x}\right)^{2}} \, dx. \end{align} Now let $x \to \frac{1}{x}$ to obtain \begin{align}\tag{2} e^{2 i a} \, I &= \sqrt{i a} \, \int_{0}^{\infty} e^{-i a \, \left( x - \frac{1}{x} \right)^{2}} \, \frac{dx}{x^{2}}. \end{align} Adding equations (1) and (2) lead to \begin{align} \frac{2 \, e^{2 i a}}{\sqrt{i a}} \, I = \int_{0}^{\infty} e^{-i a \, \left(x - \frac{1}{x}\right)^{2}} \, \left(1 + \frac{1}{x^{2}}\right) \, dx. \end{align} Let $y = x + \frac{1}{x}$ to obtain \begin{align} \frac{2 \, e^{2 i a}}{\sqrt{i a}} \, I = \int_{0}^{\infty} e^{-i a \, y^{2}} \, dy \end{align} Set $y = \sqrt{\frac{x}{i a}}$ to obtain \begin{align} \frac{2 \, e^{2 i a}}{\sqrt{i a}} \, I = \frac{1}{2} \, \sqrt{\frac{\pi}{i a}} \end{align} From this the integral is given by \begin{align} \int_{0}^{\infty} e^{- \left(u^{2} - \frac{a^{2}}{u^{2}}\right)} \, du = \frac{\sqrt{\pi}}{4} \, e^{-2 i a}. \end{align} When $4 \sqrt{t} a = \alpha$ then \begin{align} \int_{0}^{\infty} e^{- \left(u^{2} - \frac{\alpha^{2}}{16 \, t \, u^{2}}\right)} \, du = \frac{\sqrt{\pi}}{4} \, e^{- \frac{i \, \alpha}{2 \sqrt{t}}}. \end{align}

Answer 3

I assume that the integral of interest $I$ is

$$I=\int_0^{\infty}e^{-\left(u^2+\frac{\alpha^2}{16tu^2}\right)}du$$

First, we let $a^2=\frac{\alpha^2}{16t}$ and write

$$\begin{align} I&=\int_0^{\infty} e^{-a\left(\frac{u^2}{a}+\frac{a}{u^2}\right)}du\\\\ &=e^{-2a}\int_0^{\infty} e^{-a\left(\frac{u}{\sqrt{a}}-\frac{\sqrt{a}}{u}\right)^2}du\tag 1 \end{align}$$

Now, enforcing the substitution $u/\sqrt{a}\to -\sqrt{a}/u$, we find

$$I=e^{-2a}\int_0^{\infty}e^{-a\left(\frac{u}{\sqrt{a}}-\frac{\sqrt{a}}{u}\right)^2} \frac{a}{u^2}\,du \tag 2$$

Adding $(1)$ and $(2)$ and dividing by $2$ yields

$$I=e^{-2a}\int_0^{\infty}e^{-a\left(\frac{u}{\sqrt{a}}-\frac{\sqrt{a}}{u}\right)^2} \left(1+\frac{a}{u^2}\right)\,du $$

Making one final substitution $\frac{u}{\sqrt{a}}-\frac{\sqrt{a}}{u}\,\to\,u$ and we obtain

$$\bbox[5px,border:2px solid #C0A000]{I=e^{2a}\int_{-\infty}^{\infty}e^{-au^2}\sqrt{a}du=\sqrt{\pi}e^{-2a}=\sqrt{\pi}e^{-\alpha/2\sqrt{t}}}$$