Computing an improper integral

Question

I would like to compute the integral $$ I= \int_{0}^{\infty}x\lambda e^{-\lambda x}\frac{(\lambda e^{-\lambda x})^{k}-(\lambda e^{-\lambda x})^{-k}}{2k}dx $$

where $ \lambda $ is a positive real number and $ k $ is a positive integer. In doing so, I considered the integrals \begin{align} I_{1}&=\int_{0}^{\infty}x\lambda e^{-\lambda x}\frac{(\lambda e^{-\lambda x})^{k}}{2k}dx \nonumber \\ I_{2}&=\int_{0}^{\infty}x\lambda e^{-\lambda x}\frac{(\lambda e^{-\lambda x})^{-k}}{2k}dx \nonumber \end{align}

Then we have that $ I=I_{1}-I_{2} $. If $ I_{1} $ and $ I_{2} $ are both convergent, then $ I $ is convergent as well. If exactly one of them is divergent, then $ I $ will be divergent as well. Finally, if both are divergent, we can't really say anything about their difference.

Computation shows that $ I_{1} $ is convergent for such $ k $ and $ \lambda $ and its value is $ \frac{\lambda^{k-1}}{2k(k+1)^{2}} $. On the other hand, $ I_{2} $ is divergent for $ \lambda (k-1) \geq 0 $ and convergent otherwise. Since by the hypothesis, $ k $ is a positive integer and $ \lambda $ is a positive real number, we must have that $ I_{2} $ is divergent.

Summing it all up, we conclude that $ I $ is divergent.

I would appreciate any comments on my reasoning. Thank you!

Answer 1

Well, we have:

$$x\cdot\lambda\cdot e^{-\lambda x}\cdot\frac{\left(\lambda\cdot e^{-\lambda x}\right)^\text{k}-\left(\lambda\cdot e^{-\lambda x}\right)^{-\text{k}}}{2\cdot\text{k}}=x\cdot\frac{\lambda^{1+\text{k}}\cdot e^{-\lambda x\left(1+\text{k}\right)}}{2\cdot\text{k}}-x\cdot\frac{\lambda^{1-\text{k}}\cdot e^{\lambda x\left(\text{k}-1\right)}}{2\cdot\text{k}}\tag1$$

So, you need to find:

$$\frac{\lambda^{1+\text{k}}}{2\cdot\text{k}}\int_0^\infty xe^{-\lambda x\left(1+\text{k}\right)}\space\text{d}x-\frac{\lambda^{1-\text{k}}}{2\cdot\text{k}}\int_0^\infty xe^{\lambda x\left(\text{k}-1\right)}\space\text{d}x\tag2$$