Does This Succumb To Standard Techniques?

Calculus Level 3

$\large \int_{- \infty} ^ \infty \frac { e^{2x} - e^x } { x ( e^{2x}+1)( e^x+1) } \, dx$

The integral above has a closed form. Find this closed form.

Give your answer to 3 decimal places.

The answer is 0.693.

2 solutions

Jatin Yadav
Apr 21, 2014

Consider $I(a) = \displaystyle \int_{-\infty}^{\infty} \dfrac{e^{ax}-e^x}{x(e^x+1)(e^{ax}+1)} dx$

Differentiate with respect to $a$ to get:

$\dfrac{d(I(a))}{da} = \displaystyle \int_{-\infty}^{\infty} \dfrac{e^{ax}}{(e^{ax}+1)^2} dx$

Putting $e^{ax} = t$ , and solving, we get

$\dfrac{d(I(a))}{da} = \dfrac{1}{a}$

Integrating, we get

$I(a) = \ln a + c$

As $I(1) = 0$ , $c=0$ , $\Rightarrow I(a) = \ln a$

We have to find $I(2)$ . Hence, answer is $\ln 2 = 0.693$

Superb solution, I must say.

Vijay Raghavan - 7 years, 1 month ago

Could you please explain what happens to the x and 1+ exp(x) in the denominator on differentiating wrt a. Thanks

Karan Bhuwalka - 7 years, 1 month ago

can anyone tell me why the answer is not 0. if we break up the given integral into two partial fractions and then in one of them put 2x=t, then i think it will come to 0.

Shubhabrota Chakraborty - 7 years, 1 month ago

Do you mean setting the numerator as such $(e^{2x} + 1) - (e^x + 1)$ then splitting them? Then both the integrals will be divergent. Which means $\infty - \infty$ is not a valid step.

Pi Han Goh - 7 years, 1 month ago

That's the reason for my title. I was contemplating "I tried partial fractions and you shouldn't", but that would be too obvious.

Do you know of any other approaches to solve this? Jatin's solution is the only one that I know of.

Calvin Lin Staff - 7 years, 1 month ago

@Calvin Lin – My method is not as elegant as Jatin's.

I observed that given integrand is even so we have to evaluate the following:

$\displaystyle I=2\int_0^{\infty} \frac{e^{2x}-e^x}{x(1+e^x)(1+e^{2x})}\,dx$

$\displaystyle \Rightarrow I=2\int_0^{\infty} \frac{1}{x}\left(\frac{1}{1+e^x}-\frac{1}{1+e^{2x}}\right)\,dx$

I then used the following series expansions:

$\displaystyle \frac{1}{1+e^x}=\frac{e^{-x}}{1+e^{-x}}=e^{-x}\sum_{k=0}^{\infty} (-1)^k e^{-kx}=\sum_{k=0}^{\infty} (-1)^k e^{-(k+1)x}$

$\displaystyle \frac{1}{1+e^{2x}}=\sum_{k=0}^{\infty} (-1)^k e^{-2x(k+1)}$

to obtain the following:

$\displaystyle I= 2\int_0^{\infty} \frac{1}{x}\left(\sum_{k=0}^{\infty} (-1)^k e^{-x(k+1)}-\sum_{k=0}^{\infty} (-1)^k e^{-2x(k+1)}\right)\,dx$

$\displaystyle \Rightarrow I=2\sum_{k=0}^{\infty} (-1)^k\int_0^{\infty} \frac{e^{-x(k+1)}-e^{-2x(k+1)}}{x}\,dx$

Use the substitution $x(k+1)=t$ to get:

$\displaystyle I=2\sum_{k=0}^{\infty} (-1)^k \int_0^{\infty} \frac{e^{-t}-e^{-2t}}{t}\,dt=2\ln 2 \sum_{k=0}^{\infty} (-1)^k$

The sum is divergent. Sometime ago, Tunk mentioned about Grandi's series here: Another series expansion trouble?? . Using that (luckily) gave the right answer.

Pranav Arora - 7 years, 1 month ago

Yes , you are true. I did the same mistake in my first attempt to Nice integral . That one is quite a similar problem, as if you put $\ln(x+1) = t$ , you get it converted to $\displaystyle \int_{1}^{\infty} \dfrac{e^{3t} - e^{t}}{t} \text{ dt }$

jatin yadav - 7 years, 1 month ago

yaa..even i am also getting 0!

DEBARTHA MANDAL - 7 years, 1 month ago

This is awesome! After putting t=exp(ax), is it vague for me that it became 1/a. How come? Can someone explain to me? Thank you very much.

Christian Baldo - 7 years, 1 month ago

Sure. $e^{ax} = t \Rightarrow ae^{ax} dx = dt$ .

Hence, the integral changes to $\displaystyle \int_{0}^{\infty} \dfrac{1}{a} \dfrac{dt}{(1+t)^2} = \dfrac{1}{a}$

jatin yadav - 7 years, 1 month ago

Nice solution at all. I also tried but done a simple mistake..

Ayush Kumar - 3 years, 3 months ago

Nice solution, except you didn't prove that derivating under the integral was possible.

Maxence Seymat - 2 years, 2 months ago

here what about e power x

Reddy Naga Vamsi Krishna - 7 years, 1 month ago

Anastasiya Romanova
Nov 20, 2014

Denote the considered integral as $I$ , then by using partial fraction decomposition, the integral can be rewritten as $I=\int_{-\infty}^\infty \frac{1}{x}\left(\frac{e^{2x}}{e^{2x}+1}-\frac{e^{x}}{e^{x}+1}\right)\,dx$ From this, we can immediately see that the integrand is even function. Therefore $I=\int_{0}^\infty \frac{2}{x}\left(\frac{e^{2x}}{e^{2x}+1}-\frac{e^{x}}{e^{x}+1}\right)\,dx$ Now, one can use Frullani's theorem by taking $f(x)=\dfrac{e^{x}}{e^{x}+1}$ , $a=2$ , and $b=1$ . Hence, $I$ is simply equal to $ln2$ .

Does This Succumb To Standard Techniques?

The answer is 0.693.

2 solutions

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