$$\int_0^{\infty} x^{-3/2} (1 – e^{-x})\, dx$$
Evaluate the above integral with the help of Beta and Gamma functions. I'm badly stuck. I'm getting Gamma of a negative number and have no clue how to proceed further. Please help me!
For your reference, answer is $2\sqrt{\pi}$.
Best Answer
You can just use ordinary integration by part. It works because $1-e^{-x} \sim O(x)$ for small $x$ and hence $\lim\limits_{x\to0+}x^{-1/2}(1-e^{-x}) = 0$. Everything else is relatively standard.
$$\begin{align} \int_0^\infty x^{-3/2} (1-e^{-x}) dx &= \int_0^\infty (1-e^{-x}) d( -2x^{-1/2})\\ &= \big[ -2x^{-1/2} (1-e^{-x})\big]_0^\infty + 2 \int_0^\infty x^{-1/2} d(1 - e^{-x})\\ &= 2 \int_0^\infty x^{-1/2} e^{-x} dx = 2\Gamma\left(\frac12\right) = 2\sqrt{\pi} \end{align} $$ The moral of the story is when dealing with integral that contains divergence pieces in the integrand, pay attention to the grouping of individual pieces. Maintaining the right grouping avoids the headache of dealing with infinities at the end.