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.


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.

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