Seeking to evaluate $\int_{0}^{-1}\frac{e^{ax}+\frac{1}{a}xe^{a/x}-1}{x}\mathrm dx$

Question

We wish to evaluate this integral, $$I=\int_{0}^{-1}\frac{e^{ax}+\frac{1}{a}xe^{a/x}-1}{x}\mathrm dx, a\ge1$$

We try: $$I=\int_{0}^{-1}\left(\frac{e^{ax}}{x}+\frac{1}{a}e^{a/x}-\frac{1}{x}\right)\mathrm dx$$

$$I=\frac{e^{-a}-1}{a^2}+\int_{0}^{-1}\frac{\mathrm dx}{x}-\int_{0}^{-1}\frac{e^{ax}}{x}\mathrm dx$$

This integral $\int_{0}^{-1}\frac{\mathrm dx }{x}$ does not converge!

Answer 1

Nor $ \int_{0}^{-1}\frac{e^{ax}}{x}$ converges

So write: $$ I=\frac{e^{-a}-1}{a^2}+\int_{0}^{-1}(\frac{1-e^{ax}}{x})\mathrm dx $$ and solve that with limits(after you have found the antiderivative of the above expression)