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

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