Could the integral be calculated by complex analysis?

Question

The following integral seems to be related to a complex integration: $$ \int_{0}^{1}e^s\left(\frac{1}{s}-\frac{1}{s^2}\right)ds. $$ However, the upper end is $1$ instead of $+\infty$ in which case usually can be dealt with a Cauchy integral. Please give me some help or hints. Thanks a lot.

Answer 1

Use the integration formula $$\int e^x[f(x)+f'(x)] dx=e^x f(x),$$ which comes by integration by parts. Then $$\int_{0}^{1}e^s\left( \frac{1}{s}-\frac{1}{s^2} \right) ds=\frac{e^s}{s}|_{0}^{1}=-\infty$$

Answer 2

The integral is equal to $-\infty$ since $$ \int_0^1e^s({1\over s}-{1\over s^2})ds{\le \int_0^1{1\over s}-{1\over s^2}ds \\=\ln |s|+{1\over s}\Big|_0^1 \\=-\infty } $$

Answer 3

$$\int e^s\left(\frac{1}{s}-\frac{1}{s^2}\right)\,ds=\frac{e^s}{s}$$ $$\int_{\epsilon}^{1}e^s\left(\frac{1}{s}-\frac{1}{s^2}\right)\,ds=e-\frac{e^{\epsilon }}{\epsilon }$$