What is an elementary way to show that for positive integer $n$ $$ \int\frac{\sin(nx) \sin x}{1-\cos x} \,dx= x + \frac{\sin (nx)}{n} + 2 \sum_{k=1}^{n-1}\frac{\sin(kx)}{k} $$ This cropped up when trying to answer Proving $\int_0^{\pi } f(x) \, \mathrm{d}x = n\pi$ at a beginning calculus level, that is, without using contour integration or Fourier integrals or Parseval's theorem (but with clever tricks allowed). There, it would suffice to know that the integral from zero to $\pi$ is $\pi$.

I've tried to do this by induction but got nowhere (except that the base for $n=1$ is easy).

I've tried expanding as a series and matching terms but it got pretty messy.


Work backwards:

Let $$F(x)=x + \frac{\sin (nx)}{n} + 2 \sum_{k-1}^{n-1}\frac{\sin(kx)}{k}$$

Then $$F'(x)=1 + \cos (nx) + 2 \sum_{k-1}^{n-1}\cos(kx)$$

Therefore, you need to prove that $$\frac{\sin(nx) \sin x}{1-\cos x} = 1 + \cos (nx) + 2 \sum_{k-1}^{n-1}\cos(kx)$$ or equivalently that $$\sin(nx) \sin(x)= (1-\cos(x)) \left(1 + \cos (nx) + 2 \sum_{k-1}^{n-1}\cos(kx) \right)(**)$$

If there is no typo in your problem, this should be a relatively easy induction question.

Or better, if your students know complex numbers, just do the standard $$z =\cos(x)+i\sin(x)$$ and calculate $$1+2z+2z^2+..+2z^{n-1}+z^n=2(1+z+z^2+..+z^{n-1}+z^n)-1-z^n=2\frac{1-z^{n+1}}{1-z}-1-z^n \\ =\frac{2-z^{n+1}-z^n}{1-z}-1=\frac{(2-z^{n+1}-z^n)(1-\bar{z})}{(1-\cos(x))^2+\sin(x)^2}-1\\ =\frac{2-z^{n+1}-z^n-2\bar{z}+z^n+z^{n-1}}{2-2\cos(x)}-1\\ =\frac{2-z^{n+1}-2\bar{z}+z^{n-1}}{2-2\cos(x)}-1\\$$

By taking the real parts we get: $$ \left(1 + \cos (nx) + 2 \sum_{k-1}^{n-1}\cos(kx) \right)=\frac{2-\cos((n+1)x)-2\cos(x)+\cos((n-1)x)-2+2\cos(x)}{2-2\cos(x)} \\ =\frac{-\cos((n+1)x)+\cos((n-1)x)}{2-2\cos(x)}=\frac{2 \sin(nx)\sin(x)}{2-2\cos(x)} $$

which is exactly what you need to prove.

P.S. You can also try to prove $(**)$ by writing $$(1-\cos(x)) \left( \sum_{k-1}^{n-1}\cos(kx) \right)=2\sin^2(\frac{x}{2})\left( \sum_{k-1}^{n-1}\cos(kx) \right)$$ and use the fact that $$\sum_{k-1}^{n-1}\cos(kx) \sin(\frac{x}{2})$$ is telescopic.

  • $\begingroup$ I fully agree with your direction of solution. For the last part, taking the real part of $(1-z)(1+ ... z^n) will not work because the real part of a product is not the product of its real parts. $\endgroup$ – Jean Marie Feb 1 '16 at 23:35
  • $\begingroup$ @Mark Fishler There are two little typos in your formula : it is $k=1$ instead of $k-1$ and final letter $h$ should be suppressed. $\endgroup$ – Jean Marie Feb 1 '16 at 23:39
  • $\begingroup$ @JeanMarie changes made. Thanks $\endgroup$ – Mark Fischler Feb 4 '16 at 15:25
  • $\begingroup$ Yes, I will be accepting the @N.S. answer, but what I was looking for is a proof of the induction statement (the initial basis for $n=1$ is fairly trivial to prove). Your clue about telescoping is relevant. I will provide as a second answer the proof of induction, which is "relatively easy" but not immediately obvious. $\endgroup$ – Mark Fischler Feb 4 '16 at 15:25
  • $\begingroup$ @JeanMarie Good point, fixed it. $\endgroup$ – N. S. Feb 5 '16 at 16:26

To prove $$\sin(nx) \sin(x)= (1-\cos(x)) \left(1 + \cos (nx) + 2 \sum_{k=1}^{n-1}\cos(kx) \right)(**)$$ by induction, first note that the basis ($N=1)$ is easy: For $n=1$ the sum is vacuously zero, and $$\sin(1\cdot x) \sin(x)= \sin^2 x = 1-\cos^2 x = (1-\cos(x)) \left(1 + \cos (1\cdot x) \right)$$

Now assume that for some given $n$ it is true that $$ (1-\cos(x)) \left(1 + \cos (nx) + 2 \sum_{k=1}^{n-1}\cos(kx) \right) = \sin(nx) \sin(x) $$ Then The expression on the right of $(**)$, for $n+1$, is $$ (1-\cos(x)) \left(1 + \cos ((n+1)x) + 2 \sum_{k=1}^{(n+1)-1}\cos(kx) \right) = (1-\cos(x)) \left(1 + \cos (nx+x) + 2 \sum_{k=1}^{n-1}\cos(kx) + 2 \cos(nx)\right) = (1-\cos(x)) \left(1 + \cos (nx) \cos x - \sin(nx) \sin x + 2 \sum_{k=1}^{n-1}\cos(kx) + 2 \cos(nx)\right) \\=1 -\cos x + \cos (nx) \cos x -\cos(nx) \cos^2 x - \sin(nx) \sin x + \sin(nx)\sin x\cos x \\+ 2 (1-\cos x)\sum_{k=1}^{n-1}\cos(kx) + 2 \cos(nx) -2 \cos(nx) \cos x $$ Here we can rearrange terms and us $\cos^2 = 1-\sin^2$ to get $$ \left(1 + 2 (1-\cos x)\sum_{k=1}^{n-1}\cos(kx) - \sin(nx) \sin x \right) \\- \cos x + \left(\cos (nx) \cos x -2 \cos(nx) \cos x \right)-\cos(nx) (1-\sin^2 x) + 2 \cos(nx) \\+ \sin(nx)\sin x\cos x + 2 \cos(nx) \\ = \left(1 + 2 (1-\cos x)\sum_{k=1}^{n-1}\cos(kx) - \sin(nx) \sin x \right) \\- \cos x -\cos (nx) \cos x +2\cos(nx) -\cos(nx) + \cos(nx)\sin^2 x + \sin(nx)\sin x\cos x \\ =\left(1 + 2 (1-\cos x)\sum_{k=1}^{n-1}\cos(kx) - \sin(nx) \sin x \right) \\- \cos x \left(1 + \cos(nx) +\sin(nx)\sin x\right) +\cos(nx) + \cos(nx)\sin^2 x $$ Now replace $\sin(nx)\sin x$ by the expression in $(**)$, which we can do because by the indcution assumption that holds for $n$: $$(1-\cos(x)) \left(1 + \cos ((n+1)x) + 2 \sum_{k=1}^{(n+1)-1}\cos(kx) \right) = (1-\cos(x)) \left(1 + \cos (nx+x) + 2 \sum_{k=1}^{n-1}\cos(kx) + 2 \cos(nx)\right) \\ =\left(1 + 2 (1-\cos x)\sum_{k=1}^{n-1}\cos(kx) - \sin(nx) \sin x \right) \\- \cos x \left(1 + \cos(nx) +\sin(nx)\sin x\right) +\cos(nx) + \cos(nx)\sin^2 x \\ = \left(1 + 2 (1-\cos x)\sum_{k=1}^{n-1}\cos(kx) - (1-\cos x) \left(1 + \cos (nx) + 2 \sum_{k=1}^{n-1}\cos(kx) \right) \right) \\- \cos x \left[1 + \cos(nx) +\sin(nx)\sin x\right] +\cos(nx) + \cos(nx)\sin^2 x \\ = \left(1 + 2 (1-\cos x)\sum_{k=1}^{n-1}\cos(kx) - 2(1-\cos x) \sum_{k=1}^{n-1}\cos(kx) - (1-\cos x) \left(1 + \cos (nx) \right) \right) \\ -\cos x \left[1 + \cos(nx) +\sin(nx)\sin x\right] +\cos(nx) - \cos(nx)\sin^2 x \\ = \left(1 - (1-\cos x) \left(1 + \cos (nx) \right) \right) \\ - \cos x \left[1 + \cos(nx) +\sin(nx)\sin x\right] +\cos(nx) - \cos(nx)\sin^2 x \\ = - \cos nx + \cos x + \cos x \cos(nx) - \cos x - \cos x \cos(nx) \\ -\cos x \sin(nx)\sin x +\cos(nx) - \cos(nx)\sin^2 x \\ = - \cos nx +\cos(nx) + \cos x - \cos x + \cos x \cos(nx) - \cos x \cos(nx) \\ -\cos x \sin(nx) \sin x + \cos(nx)\sin^2 x \\ = \sin x \left[ \sin(nx)\cos x - \cos(nx)\sin x\right] = \sin x \sin(nx +x) = \sin((n+1)x) \sin x $$ Thus we have shown that $$\sin((n+1)x) \sin x = (1-\cos(x)) \left(1 + \cos ((n+1)x) + 2 \sum_{k=1}^{(n+1)-1}\cos(kx) \right) $$ and induction is established.

IMHO, this was not a "relatively easy" induction question, but it certainly did not use anything sophisticated at a level beyond simple trig.

  • $\begingroup$ Note -- One should look at the answer given by N.S. to see why proving this statement proves the integral. That answer logically ought to appear before this one, but the site mechanics place this one first. $\endgroup$ – Mark Fischler Feb 5 '16 at 15:00

@N.S. your solution is interesting. May I indicate a slighty more direct path for its last part?

Indeed, as you have shown, taking $z=e^{ix}$, the sum for which we want a closed form formula is the real part $R$ of:



Let us remark that this formula admits the following factorization:


Let $u=\dfrac{x}{2}$ and $Z=e^{iu}$. Thus $z=Z^2$. Therefore:


$$S=\dfrac{\ 2\cos u}{-2i\sin u}\left(-2i \sin nu\right) \ (\cos nu + i \sin nu) $$

The real part of $S$ is:

$$R=2 \sin nu \cos nu \dfrac{2 \sin u \cos u}{2 \left(\sin u \right)^2}$$

$$R=\sin 2nu \dfrac{\sin 2u}{2 (1 - \cos 2u)}=\sin nx \dfrac{\sin x}{1-\cos x}$$

as desired (the form: $\sin nx/\tan \left(\dfrac{x}{2}\right)$ being interesting as well).


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