Calculate the integral $\int_{0}^{+\infty }[e^{-(\frac{a}{x})^{2}} -e^{-(\frac{b}{x})^{2}}]dx$

Question

Calculate $$\int_{0}^{+\infty }\left[e^{-(\frac{a}{x})^{2}} -e^{-(\frac{b}{x})^{2}}\right]dx,$$ with $0<a<b$

I try to construct a inner parametric integral and change the integration order, but it doesn't work well.

Answer 1

Differentiating w.r.t. $a$ (and assuming that $a>0)$, one finds $$\frac{\partial I}{\partial a}=-2a\int_0^{\infty}\frac{e^{-a^2/x^2}dx}{x^2}=-2\int_0^{\infty}e^{-t^2}dt=-\sqrt{\pi},$$ where the 2nd step is achieved by the change of variables $x=\frac{a}{t}$. Integrating this back gives $$I=\sqrt{\pi}\left(f(b)- a\right),$$ with $f(b)$ independent of $a$. Exchanging $a\leftrightarrow b$ and repeating the procedure, we find $f(b)=b$.

Answer 2

See the following integral: $$\int_0^\infty e^{\frac{-a}{x^2}}dx$$. Now if $\frac{1}{x}=t$, then $\frac{-1}{x^2}dx=dt.$ Hence the integral gets transformed to $$-\int_0^\infty e^{-at^2}\left(\frac{-1}{t^2}\right)dt$$. Now applying byparts, we have $$\int_0^\infty e^{-at^2}\left(\frac{-1}{t^2}\right)dt=e^{-at^2}\left(\frac{1}{t}\right)\Big|_{0}^{\infty}-\int_0^\infty e^{-at^2}(-2at)\left(\frac{1}{t}\right)dt$$

$$=e^{-at^2}\left(\frac{1}{t}\right)\Big|_{0}^{\infty}+2a\int_0^\infty e^{-at^2}dt$$.

Hence $$\int_{0}^{+\infty }\left[e^{-(\frac{a}{x})^{2}} -e^{-(\frac{b}{x})^{2}}\right]dx=\left[e^{-at^2}\left(\frac{1}{t}\right)-e^{-bt^2}\left(\frac{1}{t}\right)\right]\Big|_{0}^{\infty}+2a\int_\infty^{0}e^{-at^2}dt-2b\int_\infty^{0}e^{-bt^2}dt$$

As you mentioned the problem lies when we evaluate the integral at $0$. Since at $\infty$ the square bracket thing goes to $\infty$, all that remains is its evaluation at $0$. This boils down to $$lim_{\epsilon \to 0}\frac{e^{-a\epsilon^2}-e^{-b\epsilon^2}}{\epsilon}$$. Now this is in $$\frac{0}{0}$$ form, so by L'Hospital rule, this is equivalent to $$lim_{\epsilon \to 0}\frac{e^{-a\epsilon^2}(-2a\epsilon)-e^{-b\epsilon^2}(-2b\epsilon)}{1}=0$$.

Now all that remains is the evaluation of integral on the right which you can do.

Answer 3

The difficulty arrises for $x$ tending to $\infty$. An asymptotic expension solves it.