$$\int_{0}^{\infty} \left(e^{-\frac{1}{x^2}}-e^{-\frac{4}{x^2}}\right) dx$$
I've tried much of the techniques used in the textbook, none have led to anything concrete, or i am not just able to see the answer. Can anyone type out the answer with detail if it's not a problem, it would be very appreciated. I'm in a bit of a rush.. Thankful in advance.
On
Change variables to $y=1/x^2$, so $ dx = -\frac{1}{2}y^{-3/2} \, dy $, and we have $$ \frac{1}{2}\int_0^{\infty} \frac{e^{-y}-e^{-4y}}{y^{3/2}} \, dy $$ Now, write the integrand as $$ \frac{1}{y^{1/2}}\int_1^4 e^{-ty} \, dt, $$ and interchange the order of integration: $$ \frac{1}{2}\int_1^4 \int_0^{\infty} y^{1/2-1} e^{-ty} \, dy \, dt. $$ The inner integral is well-known: it is $\Gamma(1/2) t^{-1/2} = \frac{\sqrt{\pi}}{\sqrt{t}} $, and hence the integral is $$ \sqrt{\pi} \int_1^4 \frac{dt}{2t^{1/2}} = (2-1)\sqrt{\pi} = \sqrt{\pi} $$
Assume $a>0$ and $b>0$.
You may integrate by parts and use a change of variable, giving $$ \begin{align} \int_{0}^{\infty}\left( e^{-a^2/x^2}-e^{-b^2/x^2} \right)dx&=\left.x\left( e^{-a^2/x^2}-e^{-b^2/x^2} \right)\right|_{0}^{\infty}-2\int_{0}^{\infty}\left( a^2e^{-a^2/x^2}-b^2e^{-b^2/x^2} \right)\frac{dx}{x^2}\\\\ &=-2\int_{0}^{\infty}\left( a^2e^{-a^2u^2}-b^2e^{-b^2u^2} \right)du\quad \left(u:=\frac1x\right)\\\\ &=(b-a)\sqrt{\pi } \end{align} $$ that is
where we have used the gaussian result $$ \int_{0}^{\infty}e^{-c^2u^2}du=c\sqrt{\pi },\quad c>0. $$ Then apply it to $a:=1$ and $b:=2$.