0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I think you can't integrate it by hand: wolframalpha.com/input/… $\endgroup$ – idm Aug 14 '14 at 11:28
  • $\begingroup$ 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 $\log 2$. $\endgroup$ – aziiri Aug 14 '14 at 11:41
3
$\begingroup$

This is not a solution (it is an approximation), but I hope that it leads someone to a solution.

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 $10^{-46}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.