Fast way to show that $\int_{0}^{1}x(1-x) \sin (n\pi x)dx = \frac{4}{(n\pi)^3}$, $n$ odd

Question

I can compute and show that for odd values of $n$, $$\int_{0}^{1}x(1-x) \sin (n\pi x)dx = \frac{4}{(n\pi)^3}$$ fairly easily by expanding the quadratic, splitting the integrand and then evaluating each component using integration by parts. However, this is rather tedious.

I was wondering if you could see a way of proving this identity without having to go through so much brute-force computations. The result is relatively neat so I suspect there may be a much easier way of computing the integral.

The integral looks perfect for $$\int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx$$ but I was unable to find a solution using that because I could not somehow find a way to work with $\sin (n\pi(1-x))$

So what I have so far is

$$I=\int_{0}^{1}x(1-x) \sin (n\pi x)dx=\int_{0}^{1}x(1-x) \sin (n\pi(1-x))dx$$

But I am now stuck.

So, for a general $f$, what should $$\int_0^b f(x)f(b-x)\sin (n \pi x), dx$$ be? — Siminore, Aug 18 '15 at 19:27
Do you mean that in the sense of "I actually don't know what the answer should be!" or in the cheeky sense that a teacher would say to a student sometimes? I'm tearing my hair out trying to observe some kind of symmetry property to take advantage of! — Trogdor, Aug 18 '15 at 19:41

Leucippus · Answer 1 · 2015-08-18T19:47:33.997

6

Consider the integrals \begin{align} I_{1}(n) &= \int_{0}^{1} \cos(n\pi x) \, dx = \left[ \frac{\sin(n \pi x)}{n \pi}\right]_{0}^{1} = \frac{\sin(n \pi)}{n \, \pi} \\ I_{2}(n) &= \int_{0}^{1} \sin( n \pi x) \, dx = \frac{\cos(n \pi) - 1}{n \, \pi} \end{align} Now the integral in question is obtained from: \begin{align} \int_{0}^{1} x(1-x) \, \sin( n \pi \, x) \, dx &= - \frac{1}{\pi} \, \partial_{n} I_{1} + \frac{1}{\pi^{2}} \, \partial_{n}^{2} I_{2} , \hspace{10mm} \partial_{n} = \frac{d}{dn}\\ &= - \frac{1}{\pi^{2}} \, \partial_{n}\left( \frac{\sin(n \pi)}{n} \right) + \frac{1}{\pi^{3}} \, \partial_{n}^{2} \left( \frac{1}{n} - \frac{\cos(n \pi)}{n}\right) \\ &= - \frac{1}{\pi^{2}} \left( \frac{(-1)^{n}}{n} \right) + \frac{1}{\pi^{3}} \left( \frac{2}{n^{3}} - \left(\frac{2 (-1)^{n}}{n^{3}} - \frac{\pi^{2} \, (-1)^{n}}{n} \right) \right) \\ &= \frac{2 \, (1 - (-1)^{n})}{(n \, \pi)^{3}} \end{align}

edited Aug 18 '15 at 19:47

answered Aug 18 '15 at 19:42

Leucippus

26,329

This is nice, but how general is your method? – Siminore Aug 18 '15 at 19:44
I unfortunately am not quite understanding how you acquired the first line. What does $\partial_n$ indicate? – Trogdor Aug 18 '15 at 19:44
@Trogdor The notation has been added to the solution, but in general, $\partial_{x} = \frac{d}{dx}$ for a single variable – Leucippus Aug 18 '15 at 19:48
@leucippus Feynman Lives! +1 – Mark Viola Aug 18 '15 at 21:07
@Dr.MV If I recall somewhat correctly it was one of the Bernoulli members, Euler, Gauss or Laplace that have used the Feynman trick first. In any event it can be used it often becomes handy. – Leucippus Aug 18 '15 at 22:56
Yes, of course Richard didn't "invent" the trick. But, his name seems to be associated with it. – Mark Viola Aug 18 '15 at 23:08
@Dr.MV Feynman's bag of tricks has become more wide spread. Nothing wrong with knowing a few extra tools. – Leucippus Aug 18 '15 at 23:33
And Feynman was one of the best ever. Have you seen any of his lectures? He was gifted as both a theoretical physicist and a lecturer. – Mark Viola Aug 19 '15 at 02:49

score 2 · Answer 2 · answered Aug 18 '15 at 21:38

Let $s(x)$ be the sawtooth wave, i.e. the $2\pi$-periodic function that over $(-\pi,\pi)$ equals $x$.

It is well-known that its Fourier series is given by: $$ s(x) = 2\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\,\sin(nx), $$ hence by termwise integration and evaluation at $x=0$ it follows that: $$ p(x) = 2\sum_{n\geq 1}\frac{(-1)^n}{n^2}\,\cos(nx),$$ where $p(x)$ is the "parabolic wave", i.e. the $2\pi$-periodic function that equals $\frac{x^2}{2}-\frac{\pi^2}{6}$ over $(-\pi,\pi)$.

If $g(x)$ is a $2\pi$-periodic function that equals $\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{\pi}\right)$ over $(-\pi,\pi)$, it follows that:

$$ g(x) = \frac{2}{3}-\frac{4}{\pi^2}\sum_{n\geq 1}\frac{(-1)^n}{n^2}\,\cos(nx) $$ hence: $$ h(x) = \frac{2}{3}-\frac{4}{\pi^2}\sum_{n\geq 1}\frac{1}{n^2}\,\cos(nx)$$ is the Fourier series of the $2\pi$-periodic function that equals $\frac{x}{\pi}\left(2-\frac{x}{\pi}\right)$ over $(0,2\pi)$, so:

$$ f(x) = \frac{1}{6}-\frac{1}{\pi^2}\sum_{n\geq 1}\frac{\cos(2\pi n x)}{n^2} $$ is the Fourier series of the $1$-periodic function that equals $x(1-x)$ over $(0,1)$.

That leads to a not-so-short but quite simple alternative solution.

Nice use of Fourier series where some may least expect it. – Leucippus Aug 18 '15 at 22:53 — Leucippus, Aug 18 '15 at 22:53

Fast way to show that $\int_{0}^{1}x(1-x) \sin (n\pi x)dx = \frac{4}{(n\pi)^3}$, $n$ odd

2 Answers2