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I am trying to find whether the following integral converges or diverges, but I am a bit uncertain about the steps I'm taking. I have to following integral.

$$\int_{-1}^1 \frac{e^x}{e^x-1} \,dx \\$$

Because I know it is discontinuous at x=0, I split the integral

$$\int_{-1}^0 \frac{e^x}{e^x-1} \,dx + \int_{0}^1 \frac{e^x}{e^x-1} \,dx \\$$

Now I set up the limit

$$\lim\limits_{t \to 0} \int_{-1}^t \frac{e^x}{e^x-1} \,dx + \lim\limits_{t \to 0} \int_{t}^1 \frac{e^x}{e^x-1} \,dx \\$$

After integrating using the following substitution:

$$u={e^x-1} \, \\$$

I get the following:

$$\lim\limits_{t \to 0} \int_{\frac1e-1}^{e^t-1} \frac{1}{u} \,du + \lim\limits_{t \to 0} \int_{e^t-1}^{\frac1e-1} \frac{1}{u} \,du \\$$

So finally, I get

$$\lim\limits_{t \to 0} {(ln(e^t-1)-ln(\frac1e-1))} + \lim\limits_{t \to 0} (ln(\frac1e-1)-ln(e^t-1)) \\$$

So my question is, from the last line, if t is going to 0, shouldn't e^t tend to 1 and thus I would get ln (0) which would make the integral divergent. This is not the answer in the book. Where have I gone wrong?

(Also, sorry for formatting, I'm still learning how to use all the symbols.)

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  • $\begingroup$ I don't know what is in the back of the book, but this integral is divergent $\endgroup$ – imranfat Nov 15 '16 at 22:29
  • $\begingroup$ You might be trying to find a PV integral, which looks to me like it is convergent. But as an "ordinary" improper integral, it is divergent. Which is it? $\endgroup$ – zhw. Nov 15 '16 at 22:31
  • $\begingroup$ The limits of the integral must be laterals, approaching from below in the first one, and approaching from above in the second. $\endgroup$ – Masacroso Nov 15 '16 at 22:33
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Let us study $$I=\int_0^1\frac{e^x}{e^x-1}dx$$

with the change $t=e^x$,

then $I$ becomes:

$$\int_1^e\frac{dt}{t-1}$$ which diverges

since

$$\lim_{X\to 1^+}\int_X^2\frac{dt}{t-1}=\lim_{X\to1^+}(-\ln(|X-1|))=+\infty.$$

Your integral is therefore divergent.

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  • $\begingroup$ Well the OP may be trying to compute a PV integral. $\endgroup$ – zhw. Nov 15 '16 at 22:30
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Probably tricky !

Let us consider $$I=\int_{-1}^0 \frac{e^x}{e^x-1} \,dx\qquad J= \int_{0}^1 \frac{e^x}{e^x-1} \,dx $$ For the first integral, let us substitute $x=-y$; this gives $$I=\int_{-1}^0 \frac{e^x}{e^x-1} \,dx=-\int_1^0\frac{e^{-y}}{e^{-y}-1}\,dy=\int_1^0\frac{dy}{e^y-1}=-\int_0^1\frac{dy}{e^y-1}=-\int_0^1\frac{dx}{e^x-1}$$ So $$I+J=\int_0^1 dx=1$$

We should get the same result using Pricipal Value integral.

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