# Convergence test of the following improper integral $\int_0^\infty \frac {e^{-1/x}-1} {\ x^{2/3}}dx$

I've been trying for a couple of hours to prove the convergence of the following integral: $$\int_0^\infty \frac {e^{-1/x}-1} {\ x^{2/3}}dx$$ Eventually I understood from Wolfram-Alpha that the integral converges to $\Gamma(-1/3)$ and it makes sense because by substitution $t=1/x$ I get the following integral: $$\int_0^\infty \frac {e^{-t}-1} {\ t^{4/3}}dt$$ And it's pretty close to $\Gamma(-1/3)$ except for the $-1$ that bothers me. What have I done wrong until this point? And how can I prove it? Thank you.

• There is a missing minus somewhere Commented Feb 3, 2015 at 20:47
• Actually i get: $$-\int_\infty^0 \frac {e^{-t}-1} {\ t^{4/3}}dt = \int_0^\infty \frac {e^{-t}-1} {\ t^{4/3}}dt$$ Anything done wrong here? As long as I see no minus missing. Commented Feb 3, 2015 at 21:02
• @EvgenyA. Nothing wrong. Your substitution is fine.
– J126
Commented Feb 3, 2015 at 21:02
• @EvgenyA. Here is an explanation. We let $$I(k)=\int_0^{\infty}\frac{e^{-kx}-1}{x^{\frac{4}{3}}}\, dx.$$ Then $I$ is a function of $k$. So, $$I'(k)=\int_0^{\infty}\frac{d}{dk}\frac{e^{-kx}-1}{x^{\frac{4}{3}}}\,dx=\int_0^{\infty}\frac{-xe^{-kx}}{x^\frac{4}{3}}\,dx$$ Canceling powers of $x$ gives us $$I'(k)=\int_0^{\infty}\frac{-e^{-kx}}{x^{\frac{1}{3}}}\,dx$$ Now you can do a substitution for $u=kx$ and get the integral $$I'(k)=-\frac{1}{k^{\frac{2}{3}}}\int_0^{\infty}\frac{e^{-u}}{u^\frac{1}{3}}\, du.$$
– J126
Commented Feb 4, 2015 at 11:26
• This is $$I'(k)=-\frac{1}{k^{\frac{2}{3}}}\Gamma\left(\frac{2}{3}\right).$$ Now integrate with respect to $k$: $$I(k)=\int I'(k)\, dk=-3k^\frac{1}{3}\Gamma\left(\frac{2}{3}\right).$$ Plug in $k=1$ and use the property of $Gamma$ that says $\Gamma(1+t)=t\Gamma(t)$ to get what you want.
– J126
Commented Feb 4, 2015 at 11:32

Hint: Let $I(k)=\displaystyle\int_0^\infty\frac{e^{-kx}-1}{x^n}~dx$. Now, evaluate $I'(k)$ by differentiating under the integral sign. Then integrate that expression, and let $k=1$.
• Yes. $I(k)=\displaystyle\int I'(k)~dk.~$ But $I'(k)$ has such a simple expression in terms of k... Commented Feb 3, 2015 at 23:13
• Sorry but I just don't get how it helps me here.. You mean something like that $I(k)=\displaystyle\int_0^\infty\frac{e^{-kx}-1}{x^n}~dx=\int_0^{\infty} \int_0^{\infty} -k\frac{e^{-kx}}{x^n}~dx = I'(k)$ ? I guess it's without the double integral but why is this even true then? $I(k)$ is a sequence of integrals or I completely misunderstood you hint? Commented Feb 3, 2015 at 23:32
• You're supposed to differentiate with regard to k... And there aren't any limits of integration on $\displaystyle\int I'(k)~dk$. Commented Feb 4, 2015 at 0:32
• @Lucian If $n=\dfrac{4}{3}$, don't you end up with $$\int_0^{\infty}\frac{e^{-kx}}{x^{\frac{4}{3}}}\, dx,$$ which doesn't exist?