Integral evaluation - analytical vs. numerical

[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.

 

[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..

 

[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.

 

[Zurtex]

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

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

 

[dextercioby]

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

Daniel.

EDIT:See post #7 for details.

 

[Zurtex]

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

Danie.

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?

 

[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.

 

[dextercioby]

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?

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.

 

[uart]

So just in case anyone is at all unclear about how this relates to the original observation that :

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.

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".

 

[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.

 

[krab]

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

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.

 

[broegger]

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