Improper integral: $\int^{+\infty}_0e^{-(a^2t^2+b^2/t^2)}$

Question

Consider the application $f: (\mathbb{R}^{+*})^2 \to \mathbb{R}$ given by

$f(a,b)=\int^{+\infty}_0e^{-(a^2t^2+b^2/t^2)}$

Calculate $f(a,b)$.

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I thought to take the derivative inside the integral sign (this needs justification) I obtain:

$\frac{\partial^2 f}{\partial a\partial b}=4abf$

I don't know how to solve this differential equation, but two obvious solutions

are $f=Ke^{a^2+b^2}$ and $f=Ke^{-(a^2+b^2)}$, where $K$ is a constant. The first solution clearly doesn't work, because if we increase $a$ or $b$, the integral $f(a,b)$ should decrease.

So I have a good feeling that the answer is $f(a,b)=Ke^{-(a^2+b^2)}$, but to prove it rigorously I still need to

• justify that we can take the derivative inside the integral sign

• properly solve the differential equation

• determine the constant $K$

Any hints?

Answer 1

As a special case to the Glasser's master theorem:

When $c>0$, $$\int_{-\infty}^{\infty} f(x-\frac{c}{x}) dx = \int_{-\infty}^{\infty} f(x) dx$$

Therefore $$\begin{aligned}\int_0^\infty {\exp \left( -{{a^2}{t^2} - \frac{{{b^2}}}{{{t^2}}}} \right)dt} &= \frac{1}{2}\int_{ - \infty }^\infty {\exp \left[ { - {a^2}{{(t - \frac{b}{{at}})}^2} {-2ab}} \right]dt}\\ & = \frac{1}{2}{e^{{-2ab}}}\int_{ - \infty }^\infty {\exp \left( { - {a^2}{t^2}} \right)dt} \\ & = \frac{1}{2}e^{-2ab}\frac{\sqrt{\pi}}{a} \end{aligned}$$

Answer 2

I think the variables $a$ and $b$ are positive to begin with.

The differential equation approach is valid, but not as you planned to do.

As you can see the answer by @pisco125, the form $Ke^{-a^2-b^2}$ is not the answer. This means that approaching from the following PDE $$ \frac{\partial^2}{\partial a\partial b} f = 4ab f $$ is impossible.

However, we can still approach from a differential equation. We can prove that $$ \frac{\partial}{\partial b} f = -2a f. $$ If this is true, then we integrate w.r.t. $b$ to obtain $$ f(a,b)=\exp(-2ab+g(a)). $$ To find $g(a)$, put $b=0$. This gives $$ f(a,0)=\exp(g(a))=\int_0^{\infty} \exp(-a^2 t^2 ) dt = \frac{\sqrt \pi}{2a} $$ yielding the correct answer.

For the justification of differentiation under integral sign, fix $a>0$, $b>0$ and $b_n$ be a sequence converging to $b$, not equal $b$.

Consider $$ \frac{f(a,b_n)-f(a,b)}{b_n-b}=\int_0^{\infty} \exp(-a^2 t^2) \frac{ \exp\left( - \frac{b_n^2}{t^2}\right)-\exp\left(-\frac{b^2}{t^2}\right) }{b_n-b}dt. $$ For sufficiently large $n$, $b/2\leq b_n\leq 2b$, and by Mean Value Theorem, $$ \left| \frac{ \exp\left( - \frac{b_n^2}{t^2}\right)-\exp\left(-\frac{b^2}{t^2}\right) }{b_n-b}\right|\leq \frac1{t^2}\exp\left( -\frac{(b/2)^2}{t^2}\right). $$ By Dominated Convergence Theorem, we have $$ \lim_{n\rightarrow\infty} \frac{f(a,b_n)-f(a,b)}{b_n-b}=\int_0^{\infty} \frac{-2b}{t^2} \exp(-a^2 t^2-\frac{b^2}{t^2}) dt. $$ The last integral can be shown to be equal $-2af$ by the substitution $atu=b$.