Proving the Integral for n>1: A Scientific Approach

[Daggy]

Homework Statement

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

Homework Equations

The Attempt at a Solution

Tried substitute x = cosh theta, then

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

$$\int\limits_{0}^{\infty}{\frac{1}{(x + \sqrt{x^2 + 1})^n}} \mathrm{d}x = \int\limits_{0}^{\infty}{\frac{\sinh{\theta}}{(\cosh{\theta} + \sinh{\theta})^n }$$
I'm getting in the right direction here? I'm really stuck..

 

[HallsofIvy]

Homework Statement

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

Homework Equations

The Attempt at a Solution

Isn't something missing here?

 

[Daggy]

n > 1

 

[gabbagabbahey]

I think he meant your attempt at a solution

 

[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]

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

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

$$\frac{1}{2} \int\limits_{x = 0}^{x = \infty} {e^{\theta - n\theta}} \mathrm{d}\theta - <br /> \frac{1}{2} \int\limits_{x = 0}^{x = \infty}{e^{-\theta - n\theta} \mathrm{d}\theta =$$

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 doesn't it?

 

[Dick]

Is that a problem? when integrating and substituting back the cosh dissappears doesn't 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
$$\int\limits_{0}^{\intfty}{\frac{\cosh{\theta}}{(\sinh{\theta} + \cosh{\theta})^n}}\mathrm{d}\theta = \frac{1}{2}\int\limits_{x = 0}^{x = \infty}{\frac{e^{x} + e^{-x}}{(e^x)^n}}{\mathrm{d}\theta}$$
and

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

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