# Thread: Evaluate the integral

1. ## Evaluate the integral

Evaluate f(x), defined below, in terms of x.

Hint: you may need to use special functions.

$\dpi{150} f(x) = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos(t)^x dt$

2. ## Re: Evaluate the integral

Use Gamma function.

3. ## Re: Evaluate the integral

Lol, way to spoil the solution without even giving one xD

4. ## 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

$\displaystyle s = \cos^2 t$

5. ## 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.)

6. ## Re: Evaluate the integral

Now, the big question is. Can we do it in other ways?

7. ## 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.

8. ## Re: Evaluate the integral

Spoiler:
Sorry folks! I had no clue of your original intentions...I also did not know about this SPOILER option here. Thanks for the clue.

9. ## Re: Evaluate the integral

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

Hint: you may need to use special functions.

$\dpi{150} f(x) = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos(t)^x dt$
We can use directly the property of the beta function :

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

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