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I want to compute an integral like $\int_0^{+\infty} \ln(1+x)e^{-x}\,\mathrm dx$. Then denote $\mu = 1-e^{-x}$, so $x=-\ln(1-\mu)$. Substitute this into the integral, we get

$$\int_0^1 \ln(1-\ln(1-\mu))(1-\mu)\,\mathrm d(-\ln(1-\mu)) = \int_0^1 \ln(1-\ln(1-\mu))\,\mathrm d\mu$$

However, $\ln(1-\ln(1-\mu))$ goes to infinity if $\mu$ goes to 1, which makes the latter integral uncomputable. But the original integral has no such problem. Is there any error in my substitution?

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  • $\begingroup$ The substitution is ok, as Ilya pointed out. Conceptually, you start from an improper integral as $x\rightarrow\infty$. It's indeed a limit and it's not surprising that after your substitution, you still obtain an improper integral. This time it's because your integrand is unbounded. $\endgroup$
    – user1783444
    Commented Jun 5, 2013 at 8:52
  • $\begingroup$ So you mean $\int_0^{+\infty} \ln(1+x)e^{-x}\,\mathrm dx$ is an improper integral? But when $x->+\infty$, the integrated function is 0. $\endgroup$
    – parfois
    Commented Jun 5, 2013 at 9:01
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    $\begingroup$ Yes, $\lim_{x\rightarrow\infty}\ln(1+x)e^{-x}=0$, you cannot 'really' substitute $x=\infty$. This url may be useful: en.wikipedia.org/wiki/Improper_integral $\endgroup$
    – user1783444
    Commented Jun 5, 2013 at 9:08
  • $\begingroup$ Thanks, user. I should have studied concept of improper integral. $\endgroup$
    – parfois
    Commented Jun 5, 2013 at 9:10

2 Answers 2

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You reduce the integral over the infinite interval to the one over the finite interval. You shall not be surprised that the function under the integral may become unbounded. The point is that you may think of the value of the integral as an area under the graph of the function. Now, what you change a variable, you make a correspondent change of the coordinate system where they area is defined. That shall give you an intuition why the function may (but doesn't have to) become unbounded. However, it still stays integrable.

Your operations with $\mu$ seem fine to me. To support this, I computed both integral in Mathematica (see the image below), they're both well-defined and have the same value.

For some reason, Wolfram Alpha computes only an approximate value of the second integral.

enter image description here

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  • $\begingroup$ Thanks, Ilya. But I still can't understand why there's bounded results when the function for integral is unbounded. $\endgroup$
    – parfois
    Commented Jun 5, 2013 at 8:55
  • $\begingroup$ I think I've got some point. The reason is that $\ln(1-\ln(1-\mu))d\mu$ goes to 0 as $\mu$ goes to 1. $\endgroup$
    – parfois
    Commented Jun 5, 2013 at 9:08
  • $\begingroup$ @parfois: sure, although that's neither sufficient, nor necessary. Are you familiar with improper Riemann integrals - that is either unbounded intervals, or unbounded functions on bounded intervals? E.g. when $$ \int_0^1 \frac{\mathrm dx}{x^p}, \quad \int_1^\infty \frac{\mathrm dx}{x^p} $$ are finite and infinite for $p>0$? $\endgroup$
    – SBF
    Commented Jun 5, 2013 at 9:11
  • $\begingroup$ got it, thanks a lot. $\endgroup$
    – parfois
    Commented Jun 5, 2013 at 9:18
  • $\begingroup$ @parfois: nice! $\endgroup$
    – SBF
    Commented Jun 5, 2013 at 9:19
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we have $$\int_{0}^{\infty}e^{\mu x}\ln{(\beta+x)}dx=\dfrac{1}{\mu}[\ln{\beta}-e^{\mu\beta}Ei{(-\beta\mu)}]$$

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    $\begingroup$ How does this answer the question? $\endgroup$
    – Lord_Farin
    Commented Jun 5, 2013 at 8:37
  • $\begingroup$ Thanks, but I'm still curious about where the contradiction comes from. $\endgroup$
    – parfois
    Commented Jun 5, 2013 at 8:38

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