# Evaluate the integral

• Sep 10th 2012, 03:03 PM
SworD
Evaluate the integral
Evaluate f(x), defined below, in terms of x.

Hint: you may need to use special functions.

http://latex.codecogs.com/gif.latex?...} \cos(t)^x dt
• Sep 10th 2012, 03:15 PM
MaxJasper
Re: Evaluate the integral
Use Gamma function.
• Sep 10th 2012, 03:20 PM
SworD
Re: Evaluate the integral
Lol, way to spoil the solution without even giving one xD
• Sep 10th 2012, 10:05 PM
Vlasev
Re: Evaluate the integral
And btw, I bet MaxJasper just ran it through Mathematica or something and saw that the solution is made up of Gamma functions, and wrote quickly. Here's my solution:

Spoiler:

First, we exploit the evenness of the cosine function to write

$\displaystyle \int_{-\pi/2}^{\pi/2}\cos(t)^x dt = 2\int_0^{\pi/2}\cos(t)^xdt.$

Next, we have our first substitution. Let $\displaystyle u = \cos(t)$ so that $\displaystyle du = -\sin(t) dt$. Here $\displaystyle t=\arccos(u)$ and see that

$\displaystyle 2\int_0^{\pi/2}\cos(t)^xdt = -2\int_0^1 \frac{u^xdu}{\sin(\arccos(u))}.$

Using some trig we find that

$\displaystyle \sin(\arccos(u)) = (1-u^2)^{-1/2}$

so the integral becomes

$\displaystyle 2\int_0^1 u^x (1-u^2)du.$

Our second substitution aims to obtain the Beta function (not the Gamma function as MaxJasper pointed out). Let $\displaystyle t = u^2$ so that $\displaystyle u = t^{1/2}$ and $\displaystyle du = t^{-1/2}dt/2$. We see that

$\displaystyle f(x) = \int_0^1 t^{x/2-1/2}(1-t)^{-1/2}dt$

The beta function is written as

$\displaystyle B(a,b) = \int_0^1 t^{a-1}(1-t)^{b-1} dt$

and we need to put our integral into this form. We get

$\displaystyle f(x) = \int_0^1 t^{(x/2+1/2)-1}(1-t)^{(1/2)-1}dt = B\left(\frac{1+x}{2},\frac{1}{2}\right)$

Another way we can express the Beta function is as follows

$\displaystyle B(a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$

So

$\displaystyle f(x) = \frac{\Gamma((1+x)/2)\Gamma(1/2)}{\Gamma(1+x/2)} = \frac{\sqrt{\pi}\Gamma((1+x)/2)}{\Gamma(1+x/2)}$

Of course, now that we know what the substitutions are, we can baffle everyone by using the single substitution

$\displaystyle s = \cos^2 t$
• Sep 11th 2012, 12:51 PM
SworD
Re: Evaluate the integral
Nice, that's exactly the same strategy that I had seen. (And I have to admit that I couldn't even derive it myself without some help from maple.)
• Sep 11th 2012, 09:05 PM
Vlasev
Re: Evaluate the integral
Now, the big question is. Can we do it in other ways?
• Sep 11th 2012, 09:20 PM
SworD
Re: Evaluate the integral
Well, the integral came up when I was trying to find a formula for the volume of an n-sphere.

n-sphere - Wikipedia, the free encyclopedia

These results can be proven after slight manipulations with the integral you solved. But if you prove them otherwise, the integral would immediately follow, for integer values of the variable exponent. (Actually, if $\displaystyle C_n$ is the volume of the unit n-sphere, then this is easy to show: $\displaystyle f(n) = \frac{C_n}{C_n-1}$). But I'm not sure how the writers of the article found the formulas themselves, if they didn't evaluate this integral.

But this is interesting, it explains the relation of $\displaystyle \pi$ of geometric volumes to the gamma function.
• Sep 12th 2012, 03:41 PM
MaxJasper
Re: Evaluate the integral
Spoiler:
• Jan 22nd 2013, 11:50 AM
zaidalyafey
Re: Evaluate the integral
Quote:

Originally Posted by SworD
Evaluate f(x), defined below, in terms of x.

Hint: you may need to use special functions.

http://latex.codecogs.com/gif.latex?...} \cos(t)^x dt

We can use directly the property of the beta function :

$\displaystyle \beta(x,y) = 2\int^{\frac{\pi}{2}}_0 \cos^{2x-1}(\theta)\sin^{2y-1}(\theta) \, d\theta$

$\displaystyle f(x) = 2\int_{0}^{\frac{\pi}{2}} \cos^{2(\frac{x+1}{2})-1}(t)\sin^{2(\frac{1}{2})-1}(t)\, dt= \beta(\frac{x+1}{2},\frac{1}{2})$