# Integral using the gamma function

1. Apr 17, 2017

### Mr Davis 97

1. The problem statement, all variables and given/known data
I am trying to evaluate the following integral: $\displaystyle \int^{\infty}_0 (1 - e^{-x}) x^{-\frac{3}{2}} \, dx$

2. Relevant equations

3. The attempt at a solution
When I split the above integral, I get the following $\int^{\infty}_0 x^{-\frac{3}{2}} \, dx - \Gamma (-\frac{1}{2})$ = $\int^{\infty}_0 x^{-\frac{3}{2}} \, dx + 2 \sqrt{\pi}$, but that integral doesn't converge, so I'm not sure now what I should do.

2. Apr 17, 2017

### Staff: Mentor

I haven't done it, but what about taking the power series for $e^{-x}$, splitting $x^{-\frac{3}{2}}=x^{-1}\cdot x^{-\frac{1}{2}}$ and multiplying $(1-e^{-x})x^{-1}$?

3. Apr 17, 2017

### Mr Davis 97

If I use the power series of $e^{-x}$, then what I get is that $(1-e^{-x})x^{-1} = 1 - \frac{x}{2!} + \frac{x^2}{3!} - \frac{x^3}{4!} + \cdots$, which I don't see to be useful.

Also, one question, if we find that $\int^{\infty}_0 x^{-\frac{3}{2}} \, dx$ doesn't converge, why should $\int^{\infty}_0 (1 - e^{-x}) x^{-\frac{3}{2}} \, dx$ converge at all?

4. Apr 17, 2017

### Staff: Mentor

Yes, you're right. So back to your calculation. Can you show how you ended up at $\int x^{-\frac{3}{2}}dx -\Gamma(-\frac{1}{2})$?

5. Apr 18, 2017

### Dick

The integral $\int^{\infty}_0 e^{-x} x^{-3/2} \, dx$ is not $\Gamma (-\frac{1}{2})$. It's divergent. The defining relation $\int^{\infty}_0 e^{-x} x^{n-1} dx = \Gamma(n)$ only holds for $Re(n) \gt 0$.

Last edited: Apr 18, 2017