# Area under sin^10 curve without integrating

## Homework Statement

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

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

## The Attempt at a Solution

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

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 [TeX] \dfrac{a}{ \pi } [/TeX] of the area
[Tex] \int_{0}^{pi}\sin^{10}\left(x\right) [/Tex]

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

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 [Tex]2n \pi [/Tex] 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!

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## 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

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

## 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
Homework Helper
Hmmm, very weird, if the interval to integrate the wavefunction is (0,a), the function is not normalizable, because the integral is 0!! Are you sure? I used a computational solver and it agrees with the solution by hand, 63a/256.

dextercioby
Homework Helper
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.

Last edited:
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
Homework Helper
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

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

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