Integral

[Daggy]

1. The problem statement, all variables and given/known data
Prove that for n> 1

\int\limits_{0}^{\infty}{\frac{1}{(x + \sqrt{x^2 + 1})^n}} \mathrm{d}x = \frac{n}{n^2 - 1}

2. Relevant equations



3. The attempt at a solution

Tried substitute x = cosh theta, then


\frac{\mathrm{dx}}{\mathrm{d}\theta} = \sinh \theta



\int\limits_{0}^{\infty}{\frac{1}{(x + \sqrt{x^2 + 1})^n}} \mathrm{d}x = \int\limits_{0}^{\infty}{\frac{\sinh{\theta}}{(\co sh{\theta} + \sinh{\theta})^n }

I'm getting in the right direction here? I'm really stuck..

[HallsofIvy]

1. The problem statement, all variables and given/known data
Prove that

\int\limits_{0}^{\infty}{\frac{1}{(x + \sqrt{x^2 + 1})^n}} \mathrm{d}x = \frac{n}{n^2 - 1}

2. Relevant equations



3. The attempt at a solution
Isn't something missing here?

[Daggy]

n > 1

[gabbagabbahey]

I think he meant your attempt at a solution :wink:

[Daggy]

well I believe that mentioning n > 1 also is quite important... I've tried to use sinh-substitution, but can't really say I'm getting any wiser about it.

[Dick]

well I believe that mentioning n > 1 also is quite important... I've tried to use sinh-substitution, but can't really say I'm getting any wiser about it.

Use cosh(x)=(e^x+e^(-x))/2 and sinh(x)=(e^x-e^(-x))/2. Now try to integrate it. It's really pretty easy.

[Daggy]

\int\limits_{0}^{\infty}{\frac{1}{(x + \sqrt{x^2 + 1})^n}} \mathrm{d}x = \int\limits_{x = 0}^{x = \infty}{\frac{\sinh{\theta}}{(\cosh{\theta} + \sinh{\theta})^n } \mathrm{d}\theta




= \int\limits_{x = 0}^{x = \infty} \frac{e^{\theta} - e^{-\theta}}{(e^\theta)^n} \mathrm{d}\theta

Don't know what to do here.

[Dick]

Split it into two integrals. Use rules of exponents. And you are missing a factor of 1/2, I think.

[Daggy]

\frac{1}{2} \int\limits_{x = 0}^{x = \infty} {e^{\theta - n\theta}} \mathrm{d}\theta -
\frac{1}{2} \int\limits_{x = 0}^{x = \infty}{e^{-\theta - n\theta} \mathrm{d}\theta =




Then integrating and substitute back, don't really seem to get rid of those exponentials..

[Dick]

Just continue. Write down the antiderivatives. n>1. The contribution from infinity vanishes. It's the theta=0 part that counts. And e^0=1. See, no exponential?

[HallsofIvy]

Your original substitution x= cosh(theta) has a fundamental problem: the lower limit of integration is x= 0 and cosh(theta) is never 0. I would suggest x= tan(y) instead.

[Daggy]

Is that a problem? when integrating and substituting back the cosh dissappears does'nt it?

[Dick]

Is that a problem? when integrating and substituting back the cosh dissappears does'nt it?

Yeah, it's a problem. If you use x=cosh(theta), 1+cosh(theta)^2 isn't equal to sinh(theta)^2. It's the other way around. Use x=sinh(theta). (You had said you were using sinh-substitution, and I didn't notice you had it backwards.) It looks almost the same as x=cosh(theta) except for a sign difference.

[Daggy]

So

\int\limits_{0}^{\intfty}{\frac{\cosh{\theta}}{(\s inh{\theta} + \cosh{\theta})^n}}\mathrm{d}\theta = \frac{1}{2}\int\limits_{x = 0}^{x = \infty}{\frac{e^{x} + e^{-x}}{(e^x)^n}}{\mathrm{d}\theta}

and


\frac{1}{2}\int\limits_{x = 0}^{x = \infty}{e^{\theta - n\theta}}\mathrm{d}\theta + \frac{1}{2}\int\limits_{x = 0}^{x = \infty}{e^{-\theta - n\theta}}\mathrm{d}\theta



and then factorize e^{1 - n} out?

\frac{e^{1 - n}}{2} \int\limits_{x = 0}^{x = \infty}{cosh \theta}~ \mathrm{d}\theta

[Dick]

Noooo. And you can change your limits to theta=0 etc, ok? Your first integral is e^((1-n)*theta). That's EASY to integrate.

[Daggy]

ahh. figured out, thanks