# Why this integral by parts not work?

Suppose we have the original integral \begin{equation} \int_{0}^{\infty}\ln(t+a)e^{-t}e^{-e^{-t}}dt, \end{equation} 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

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