# Evaluation of Integral $\int_{-100}^{100}\frac{\left(e^{2x}-e^x\right)}{x\cdot \left(e^{2x}+1\right)\cdot \left(e^x+1\right)}dx$

Evaluation of Integral $$\int_{-100}^{100}\frac{\left(e^{2x}-e^x\right)}{x\cdot \left(e^{2x}+1\right)\cdot \left(e^x+1\right)}dx$$

My Try:: Let $\displaystyle I = \int_{-100}^{100}\frac{\left(e^{2x}-e^x\right)}{x\cdot \left(e^{2x}+1\right)\cdot \left(e^x+1\right)}dx\tag1$

Using $$\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx$$

And Let $f(x) = \frac{\left(e^{2x}-e^x\right)}{x\cdot \left(e^{2x}+1\right)\cdot \left(e^x+1\right)}\;$ So $\;\;\displaystyle f(-x)=f(x)$

So $f(x)$ is an even

So $\displaystyle I =2\int_{0}^{100}\frac{\left(e^{2x}-e^x\right)}{x\cdot \left(e^{2x}+1\right)\cdot \left(e^x+1\right)}dx$

Now How can I solve after that?

Help me

Thanks.

• I think you can't integrate it by hand: wolframalpha.com/input/… – idm Aug 14 '14 at 11:28
• Numerically it seems to be very close to $\log \sqrt{2}$ but I don't think it is the answer, maybe the integral over $\mathbb{R}$ is $\log 2$. – aziiri Aug 14 '14 at 11:41

For $a>0$ set : $$f(a)=\int_{-\infty}^{\infty} \frac{e^{ax}-e^x}{(e^{ax}+1)(e^x+1)x} \ \mathrm{d}x$$ it is differentiable and $$f'(a)= \int_0^{\infty} \frac{e^{a x}}{\left(e^{a x}+1\right)^2}\ \mathrm{d}x = \frac{1}{a} .$$Since $f(1)=0$ we have $f(a)=\ln a$ then $$0<\ln 2 - \int_0^{100} \frac{e^{2x}-e^x}{(e^{2x}+1)(e^x+1)x} \ \mathrm{d}x < \int_{100}^{\infty} \frac{e^{2x}-e^x}{100 (e^{2x}+1)(e^x+1)} \ \mathrm{d}x =\frac{1}{200} \ln \frac{(1+e^{100})^2}{1+e^{200}}$$ The right expression is about $10^{-46}$.