# Why this integral by parts not work?

Suppose we have the original integral $$\int_{0}^{\infty}\ln(t+a)e^{-t}e^{-e^{-t}}dt,$$ where $a$ is a positive constant. With the integral by parts, we can rewrite the above integral as \begin{align} &\int_{0}^{\infty}\ln(t+a)de^{-e^{-t}}\\ &=\ln(t+a)e^{-e^{-t}}|_{0}^{\infty}-\int_{0}^{\infty}\frac{e^{-e^{-t}}}{t+a}dt. \end{align} Obviously, the first term is infinite, which means that we cannot rewrite the origital integral with the integral by parts, but WHY?

• The $\int_{0}^{\infty}\frac{e^{-e^{-t}}}{t+a}dt$ diverge, so we really have an indeterminate form $\infty-\infty$, but I don't see how to calculate the integral.... Nov 11 '16 at 10:54
• i would add $0=1-1$ in your last integral (to be more formal you should replace all upper limits by $L$) . this will produce a convergent term plus something which has exactly the same typ of singular behaviour as your first term. this will cancel the artifical divergence and you can take the limit $L \rightarrow \infty$ Nov 11 '16 at 11:07
• the correct result then reads $$\int_{0}^{\infty}\log(t+a)e^{-t}e^{-e^{-t}}dt=-\int_{0}^{\infty}\frac{e^{-e^{-t}}-1}{t+a}$$ which converges superfast Nov 11 '16 at 11:13
• @tired: Thank you very much for your interesting comment. But it is not very clear to me. Could you please describe it in a little more detail? Thanks again!! Nov 11 '16 at 12:39
• @tired: Thank you, you are really professional! BTW the offset should be $\log(a)(1-e^{-1})$. Nov 11 '16 at 13:39

Your method seems fine at first glance, but try re-writing the improper integral as the following (Improper integrals cannot be computed using a normal Riemann integral.)

$$\int_{0}^{\infty}\ln(t+a)e^{-t}e^{-e^{-t}}dt = \lim_{L\to\infty}\int_{0}^{L}\ln(t+a)e^{-t}e^{-e^{-t}}dt$$

Then carry on as you are and place the limit in at the end of the evaluation.

• Thanks for your information. But it seems not very helpful if using your described method, because we cannot further simply the integral form after using the integral by parts. Nov 11 '16 at 10:44
• @kawofengche. The fact that you cannot simplify the second integral does not change the problem. Nov 11 '16 at 10:53