Area under sin^10 curve without integrating

[ephedyn]

Homework Statement

The actual question is asking for the normalization constant for the wavefunction
$$\psi\left(x\right)=A\sin^{5}\left(\dfrac{\pi x}{a}\right)$$
without carrying out integration

In short they want me to find the value of A such that
$$A^{2}\int_{0}^{a}\sin^{10}\left(\dfrac{\pi x}{a}\right)=1$$

Homework Equations

The Attempt at a Solution

I know the integral comes out nicely to $$63a/256$$ so $$\sqrt{\dfrac{256}{63a}}$$

I figure that there must be some linear map from the width of the 'half cycle' to the area under the graph. So we are taking $$\dfrac{a}{ \pi }$$ of the area
$$\int_{0}^{pi}\sin^{10}\left(x\right)$$

What remains is for me to actually show that
$$\int_{0}^{pi}\sin^{10}\left(x\right)=\dfrac{63 \pi}{256}$$

Is there some method of series expansion here I can use to show this? I think something's going to happen to take out all 2nd, 3rd, 4th... terms of the series expansion because of the $$2n \pi$$ cycles or something along those lines.

Or is there an entirely different way to find the area under the curve without integrating that I missed out? Thanks for your help!

 

[ephedyn]

Homework Statement

The actual question is asking for the normalization constant for the wavefunction

$$\psi\left(x\right)=A\sin^{5}\left(\dfrac{\pi x}{a}\right)$$

without carrying out integration. In short they want me to find the value of A such that

$$A^{2} \int_{0}^{a} \sin^{10} \left( \dfrac{ \pi x}{a} \right)=1$$

Homework Equations

The Attempt at a Solution

I know the integral comes out nicely to $$63a/256$$ so $$A=\sqrt{\dfrac{256}{63a}}$$

I figure that there must be some linear map from the width of the 'half cycle' to the area under the graph. So we are taking $$\dfrac{a}{ \pi }$$ of the area
$$\int_{0}^{\pi}\sin^{10}\left(x\right)$$

What remains is for me to actually show that
$$\int_{0}^{\pi}\sin^{10}\left(x\right)=\dfrac{63 \pi}{256}$$

Is there some method of series expansion here I can use to show this? I think something's going to happen to take out all 2nd, 3rd, 4th... terms of the series expansion because of the $$2n \pi$$ cycles or something along those lines.

Or is there an entirely different way to find the area under the curve without integrating that I missed out? Thanks for your help!

Strangely the LaTeX is not showing :/ Basically I'm asking, how do I find the area under sin^10(pi*x/a) from x=0 to x=a without actually integrating?

 

[dextercioby]

Hmmm, very weird, if the interval to integrate the wavefunction is (0,a), the function is not normalizable, because the integral is 0!

 

[ephedyn]

Are you sure? I used a computational solver and it agrees with the solution by hand, 63a/256.

 

[dextercioby]

Apologies, I have made a mistake; please, disregard my comment above. I don't know a method to <guess> the solution without computing the integral below

$$\int\limits_{0}^{1} \sin^{10}\pi x \, dx = \frac{1}{\pi} \int\limits_{0}^{\pi} \sin^{10} x \, dx$$

Since the hand computation looks tedious, perhaps the author wants you to look it up in an integral table like Gradshteyn and Rytzhik.

 

[ephedyn]

bigubau: Wow that is a GREAT book. Thanks a lot! I managed to get it... in the section on 'representation of powers of trigonometric and hyperbolic functions in terms of functions of multiples of the argument' there's a way to rewrite sin^(2n)(x) as 1/2^(2n) of {a series of 5 terms (from k=0 to 4) + 10 choose 5}. Since the terms in the series contain cos2(5-k)x, these cancel out when evaluated at 0 and pi. What remains of the definite integral is to integrate 1/2^(2n) * (10 choose 5) from 0 to pi, which gives 1/2^(2n) * (10 choose 5)pi = 63pi/256.

Nice! I owe you one ;)

 

[dextercioby]

Don't mention it. There's a trick to compute the following integral in the general case

$$I = \int\limits_{0}^{\pi} \sin^n x \, dx$$

First, use the fact that the function is symmetric around $\frac{\pi}{2}$ so that

$$I = \int\limits_{0}^{\pi} \sin^n x \, dx = 2\int\limits_{0}^{\frac{\pi}{2}} \sin^n x \, dx = B\left(\frac{n+1}{2},\frac{1}{2}\right) = \frac{\Gamma\left(\frac{n+1}{2}\right) \Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{n}{2} +1\right)}$$

$B \, \mbox{and} \, \Gamma$ are Euler's functions.

Now plug n=10 and find

$$I=\frac{\Gamma\left(\frac{11}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(6)} = \sqrt{\pi}\frac{\Gamma\left(\frac{1}{2} +5\right)}{5!} = \sqrt{\pi}^{2} \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} \cdot \frac{1}{4^5} = \pi \frac{7 \cdot 9 \cdot 4}{4^5} = \pi \frac{63}{256}$$

 

[ephedyn]

^Ah, looks more elegant. I'll keep that in mind too when I find something like this again next time.