# Integral evaluation

1. Dec 9, 2014

### nuuskur

1. The problem statement, all variables and given/known data
$$\int_{-\pi}^{\pi}x^{2014}\sin x {\rm{d}}x$$

2. Relevant equations

3. The attempt at a solution
For such problems there are probably some extremely clever solutions, but I can't see any easy way here.
If I were to find the antiderivative of this: I would eventually come up with a sum: $$-\frac{2014!}{0!}x^{0} \sin x + \frac{2014!}{1!}x\cos x + \frac{2014!}{2!}x^2 \sin x -\frac{2014!}{3!}x^3 \cos x + ... +\\ \frac{2014!}{2013!}x^{2013} \sin x -\frac{2014!}{2014!}x^{2014} \cos x$$
[is there a way to write this sum in sigma notation?]
Are there 2015 summands?
All summands with sine in them sum up to 0, because $\sin\pm\pi = 0$. and $\cos\pm\pi = -1$.
I have some series left from evaluating at pi and I have exactly the same series evaluated at -pi. Is the integral equal to 0? EDIT: NO, I forgot the x terms.

Is there any, more elegant solution to this problem, though?

Last edited: Dec 9, 2014
2. Dec 9, 2014

### Zondrina

An integral of the form $\int x^n sin(x) \space dx$ would usually be handled with multiple applications of integration by parts, but when $n$ is large this is obviously out of the question without the aid of a computer.

There is actually a standard form for such an integral, which relies on the notion of summing up the $sin(x)$ and $cos(x)$ terms (you have already noticed this). The standard form in terms of the gamma function is:

$$\int x^n sin(x) \space dx = - \frac{i^n}{2} \left[ \Gamma(n+1, -ix) - (-1)^n \Gamma(n+1, -ix) \right]$$

Substituting your limits should help you confirm.

EDIT: Alternatively, express $sin(x)$ in terms of its power series, bring the $x^{2014}$ in and presto I believe.

Last edited: Dec 9, 2014
3. Dec 9, 2014

### Dick

x^2014 is an even function. sin(x) is an odd function. What kind of function is the product?

4. Dec 9, 2014

### nuuskur

I'm not sure, going with my gut I would say it's an odd function, which means that F(-x) + F(x) = 0
The evaluation would then look like F(pi) - F(-pi) = 2F(pi)?

5. Dec 9, 2014

### Dick

Yes, it's an odd function. The integral of any odd function on an interval like [-a,a] that's symmetric around the origin is zero. You haven't seen this before?

6. Dec 9, 2014

### Zondrina

Not quite. If $f(x)$ is odd on $[-L, L]$ or $(-L, L)$, then $\int_{-L}^{L} f(x) \space dx = 0$.

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