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I met this integral: $$ \int_0^{\infty} \left(\frac{a}{(e^{ax}-1)}-\frac{b}{(e^{bx}-1)}\right)\mathrm{d}x, \text{ where } \,\,0<a<b . $$ It is a problem that showed up on my final exam of the lesson Mathematics I. Would anyone help me calculate this integral? Thanks.

cqfd
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unicornki
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  • Two ideas : maybe it's possible to use Frulanni's integrals ? And what about to represent $\frac{1}{e^{ax} - 1} = \sum $ and integrate under the sum sign ? – openspace Jan 10 '19 at 12:21
  • I think it must be zero – 28ADY0901 Jan 10 '19 at 12:25
  • For the left side make the substitution $u=e^{ax}-1$ and for the right side make the substitution $v=e^{bx}-1$ and see where this leads you – Henry Lee Jan 10 '19 at 12:42

3 Answers3

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Here is an hint for solving $\displaystyle\int \frac{a}{e^{ax}-1}dx $.

Let $e^{ax}-1=u $. This substitution will give us the integral $$\int\frac{1}{u(u+1)}du=\int\left(\frac1u-\frac{1}{(u+1)}\right)du.$$ Can you take it from here?

cqfd
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Hints: 0) Prove the integral to be absolutely convergent.

1) Integrate from $\epsilon >0$ instead and prove that you may split the integral.

2) Change variables so that integrands are the same.

3) See what happens and remember $e^x=1+x+o(x)$ at $x \rightarrow 0$.

Aphelli
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Another hint for solving $\int \frac{a}{e^{ax}-1} dx$ $$\int \frac{a}{e^{ax}-1} dx = \int \frac{a(e^{ax}+1)}{e^{2ax}-1} dx = \frac{1}{2}\int \frac{a}{e^{ax}-1} dx + \int \frac{a e^{ax}}{e^{2ax}-1} dx$$

Reagrouping terms and a change of variable s.t. $u = e^{ax}$ and $du = a e^{ax}dx$, leads to: $$\frac{1}{2}\int \frac{a}{e^{ax}-1} dx = \int \frac{1}{u^2-1} du= \frac{1}{2} \left( \int \frac{1}{u-1} du - \int \frac{1}{u+1} du\right)$$

$$ \boxed{ \int \frac{a}{e^{ax}-1} dx = \ln\left( e^{2ax} - 1\right) - \ln\left( e^{2ax} + 1\right) }$$

Carlos Campos
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