Evaluate this integral $\int_{0}^{+\infty}\frac{M}{a}x \big(1-\exp(-\frac{x}{a})\big)^{M-1}\exp(-\frac{x}{a})dx$

Question

I'm trying to evaluate this integral: $\int_{0}^{+\infty}\frac{M}{a}x \big(1-\exp(-\frac{x}{a})\big)^{M-1}\exp(-\frac{x}{a})dx$

I tried the following:

Using: $u = \frac{x}{a}$. Thus, we have:

$Ma\int_{0}^{+\infty} u \big(1-\exp(-u)\big)^{M-1}\exp(-u)du$

We can rewrite: $(1-exp(-u))^{M-1} = \sum_{i=0}^{M-1}{{M-1}\choose{i}}(-1)^ie^{-ui}$. In this manner:

$Ma\int_{0}^{+\infty} \sum_{i=0}^{M-1}{{M-1}\choose{i}}(-1)^iue^{-u(i+1)}du = \sum_{i=0}^{M-1}{{M-1}\choose{i}}(-1)^i \frac{1}{(i+1)^2} = \sum_{k=1}^{M}{{M}\choose{k-1}}(-1)^{k+1}\frac{1}{k^2}$

I'm stuck in the last step, somehow this sum is equal to harmonic serie. I tried to separate the sum in terms of odd and even members, but I didnt get nothing.

Thanks!

Answer 1

Enforce the substitution $y=e^{-x/a}$. Then, we find that

$$\frac Ma\int_0^\infty x(1-e^{-x/a})^{M-1}\,e^{-x/a}\,dx=-aM\int_{0}^1 (1-y)^{M-1}\log(y)\,dy\tag1$$

Next, substituting $z=1-y$ reveals

$$\begin{align} \frac Ma\int_0^\infty x(1-e^{-x/a})^{M-1}\,e^{-x/a}\,dx&=-aM\int_0^1 z^{M-1}\log(1-z)\,dz\\\\ &=aM\sum_{n=1}^\infty \frac1n \int_0^1 z^{M+n-1}\,dz\\\\ &=aM\sum_{n=1}^\infty \frac{1}{n(n+M)}\\\\ &=a\sum_{n=1}^\infty \left(\frac{1}{n}-\frac1{n+M}\right)\\\\ &=a\sum_{n=1}^M\frac1n\\\\ &=aH_M \end{align}$$