1
$\begingroup$

Study the convergence of the integral: $$I_{\alpha }=\int _0^{\infty }\left(\frac{e^{-\alpha x}}{\sqrt{x}}\right)\:dx$$ and calculate $I_2$.

Ok so to study the convergence I'm using convergence criterion for improper integrals. I see that this is a type III improper integral so I split it into two, like this: $$\int _0^{\infty }\left(\frac{e^{-\alpha x}}{\sqrt{x}}\right)\:dx\:=\int _0^{1\:}\left(\frac{e^{-\alpha \:x}}{\sqrt{x}}\right)\:dx+\int _1^{\infty \:}\left(\frac{e^{-\alpha \:x}}{\sqrt{x}}\right)\:dx$$

Then I am left with a type II and a type I improper integral.

For the first I calculate $$\lambda =lim_{x\rightarrow 0}\left(x^p\left(\frac{e^{-\alpha \:x}}{\sqrt{x}}\right)\:\right)$$ which I can see that for $x=1/2$ will give me $\lambda =1$. So I have $p\lt1$ and $\lambda \lt \infty$ which means it converges.

Now for the second integral, it's a type I integral. So I calculate $$\lambda =lim_{x\rightarrow \infty }\left(x^p\left(\frac{e^{-\alpha \:x}}{\sqrt{x}}\right)\:\right)$$ which for $p=2$ will give me $\lambda=0$. So I have $p\gt1$ and $\lambda \lt \infty$ so that means it also converges.

I'm not sure if everything I did is right so far, because for my second integral, the limit $\lambda$ will also be $0$ for a value of $p$ lower than $1$($p=1/2$). Also, the $\alpha$ doesn't play any role in the convergence of the integral, at least not for how I did it, so I'm assuming I did something wrong.

And I've no clue on how to solve this $$I_2=\int _0^{\infty \:\:}\left(\frac{e^{-2\:\:x}}{\sqrt{x}}\right)\:dx$$ which is the second part of the exercise.

$\endgroup$
2
$\begingroup$

What you have done seems Ok. It is important that $\alpha>0$ to ensure the convergence of the integral.

Concerning the closed form of the integral, you may perform a change of variable $$ u=\sqrt{x},\quad du=\frac1{2\sqrt{x}}dx, $$ obtaining $$ I_{\alpha }=\int _0^{\infty }\left(\frac{e^{-\alpha x}}{\sqrt{x}}\right)\:dx=2\int _0^{\infty }e^{-\alpha u^2}\:du=\frac{\sqrt{\pi }}{2 \sqrt{\alpha}}, \quad \alpha>0, $$ where we have used a celebrated result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.