- Thread starter
- #1

- Jan 17, 2013

- 1,667

$\displaystyle \int^{\infty}_0 \, \frac{\log (1+e^{ax})}{1+e^{bx}}\, dx $

I am not sure whether it can be solved

- Thread starter ZaidAlyafey
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- Thread starter
- #1

- Jan 17, 2013

- 1,667

$\displaystyle \int^{\infty}_0 \, \frac{\log (1+e^{ax})}{1+e^{bx}}\, dx $

I am not sure whether it can be solved

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- #2

- Jan 26, 2012

- 4,197

1. Let $u=bx$. Let's assume $b>0$. Then you have $du=b\,dx$, and the integral becomes

$$ \frac{1}{b} \int_{0}^{ \infty} \frac{ \ln(1+e^{au/b})}{1+e^{u}}\,du.$$

Next, you can introduce symmetry where there isn't by factoring out, up top, a $e^{au/(2b)}$, which gets you

$$ \int= \frac{1}{b} \int_{0}^{ \infty} \frac{ \ln(e^{au/(2b)}(e^{-au/(2b)}+e^{au/(2b)}))}{e^{u/2}(e^{-u/2}+e^{u/2})}\,du

= \frac{1}{b} \int_{0}^{ \infty} \frac{ (au/(2b))+\ln(e^{-au/(2b)}+e^{au/(2b)})}{e^{u/2}(e^{-u/2}+e^{u/2})}\,du$$

$$=\frac{1}{2b} \int_{0}^{ \infty} \frac{ (au/(2b))+ \ln(2 \cosh(au/(2b)))}{e^{u/2} \cosh(u/2)}\,du=\frac{1}{2b} \int_{0}^{ \infty} \frac{ (au/(2b))+ \ln(2)+ \ln( \cosh(au/(2b)))}{e^{u/2} \cosh(u/2)}\,du.$$

You could break that up into three integrals. The middle one is tractable, actually. The outer two are still problematic.

That's about as far as I can go. Perhaps someone else has other ideas? Or could run with these?

Naturally, if you know what $a$ and $b$ are, you could integrate numerically. I think the integrals will likely converge, as the original denominator will dominate the numerator significantly.

- Thread starter
- #3

- Jan 17, 2013

- 1,667

$\displaystyle \int^1_0 \, \frac{\log \left(t^{\frac{a}{b}}+1 \right)}{t+1}\, dt$

A power-series expansion looked for the first glance the first way to go ...