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!