# Integral evaluation - analytical vs. numerical

1. May 19, 2005

### broegger

Hi,

Does anyone know a reason why $$\int_{-\infty}^{\infty}\cosh(x)^{-n}dx$$ (n>0) can be evaluated analytically when n = 1,2,3,..., but only numerically when n is non-integer. I don't know if there is a "reason", but I'm using this result in a Quantum Mechanics project and it would be cool if I could give some kind of intuitive reason why this is so.

2. May 19, 2005

### arildno

My best guess:
You may use the residue theorem from complex analysis to gain an exact expression in the case of integer values, whereas this won't work if you have a non-integer.
I might be wrong, though..

3. May 19, 2005

### dextercioby

I've attached the antiderivative and here's your integral

$$\int_{-\infty}^{+\infty} \frac{dx}{\cosh^{n} x}=\frac{\sqrt{\pi}}{2}\frac{\Gamma \left(\frac{n}{2}\right)}{\Gamma \left(\frac{n+1}{2}\right)}$$

"n" can be complex,even.

Daniel.

EDIT:The correct result is multiplied by 2.See posts #4 & #7.

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Last edited: May 19, 2005
4. May 19, 2005

### Zurtex

O.K, in Mathematica I go that if $\Re (n) > 0$ then:

$$\int_{-\infty}^{+\infty} \cosh^{-n} (x) dx = \frac{ \sqrt{\pi} \, \Gamma\left( \frac{n}{2} \right)}{ \Gamma\left( \frac{n + 1}{2} \right)}$$

5. May 19, 2005

### dextercioby

How do you explain the "1/2" factor discrepancy...?

Daniel.

EDIT:See post #7 for details.

Last edited: May 19, 2005
6. May 19, 2005

### Zurtex

Well if I left n=1 and integrate in mathematica I get $\pi$ and:

$$\Gamma \left( \frac{1}{2} \right) = \sqrt{\pi}$$

Are you sure you did not integrate between 0 and Infinity?

Last edited: May 19, 2005
7. May 19, 2005

### dextercioby

I got it,i plugged only for half a domain

Using [1]

$$\int_{0}^{+\infty} \frac{\sinh^{\mu}x}{\cosh^{\nu}x} \ dx =\frac{1}{2}B\left(\frac{\mu +1}{2},\frac{\nu-\mu}{2}\right)$$ (1)

,provided that

$$\mbox{Re} \left(\mu\right) >-1 \ ; \ \mbox{Re} \left(\mu-\nu\right) <0$$ (2)

Making $\mu=0$ (3) in (1) (admissible according to (2) and implying the condition $\mbox{Re} (\nu) >0$ (4) ),one gets

$$\int_{0}^{\infty} \frac{dx}{\cosh^{\nu} x} \ dx =\frac{1}{2} B\left(\frac{1}{2},\frac{\nu}{2}\right)=\frac{1}{2}\frac{\Gamma \left(\frac{1}{2}\right)\Gamma \left(\frac{\nu}{2}\right)}{\Gamma\left(\frac{\nu+1}{2}\right)}$$ (5)

Using

$$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$$ (6)

and

$$\int_{\mathbb{R}} \frac{dx}{\cosh^{\nu} x} \ dx =2 \int_{0}^{\infty} \frac{dx}{\cosh^{\nu} x} \ dx$$ (7)

,one gets exactly the formula posted by Zurtex.

Daniel.

--------------------------------------------------
[1]G & R,5-th edition,Academic Press,CD version,1996,page 388,formula 3.512-2.

8. May 19, 2005

### dextercioby

Yes,i used an online integrator and indeed integrated only half of the domain.

I'll edit and make a reference to your post & mine just above.

Daniel.

9. May 22, 2005

### uart

So just in case anyone is at all unclear about how this relates to the original observation that :
It's because $$\Gamma(x)$$ is easily expressable in terms of elementary functions and constants for whole and half integers but generally needs other forms of numerical calulation for arbitrary values of "x".

Last edited: May 22, 2005
10. May 22, 2005

### dextercioby

Incidentally,both beta & gamma-Euler are tabulated for a certain range of values of their arguments.

So the integral is solvable analytically.Expressing its values in terms of analytical special functions means just that.

Daniel.

11. May 22, 2005

### krab

That means that whether an integral has an analytical solution or not is based on convention (how many special functions are named). I don't find such a distinction very useful. I would echo uart's comment.

12. May 25, 2005

### broegger

Thanks y'all. I've included a bit about the gamma function, interesting stuff...