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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?

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    $\begingroup$ 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.... $\endgroup$ Nov 11 '16 at 10:54
  • $\begingroup$ 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 $L \rightarrow \infty$ $\endgroup$
    – tired
    Nov 11 '16 at 11:07
  • $\begingroup$ 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 $\endgroup$
    – tired
    Nov 11 '16 at 11:13
  • $\begingroup$ @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!! $\endgroup$ Nov 11 '16 at 12:39
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    $\begingroup$ @tired: Thank you, you are really professional! BTW the offset should be $\log(a)(1-e^{-1})$. $\endgroup$ Nov 11 '16 at 13:39
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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.

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  • $\begingroup$ 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. $\endgroup$ Nov 11 '16 at 10:44
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    $\begingroup$ @kawofengche. The fact that you cannot simplify the second integral does not change the problem. $\endgroup$ Nov 11 '16 at 10:53

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