Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How would you compute the following integrals?

$$ I_{n} = \int_0^\pi \frac{1-\cos nx}{1- \cos x} dx $$ $$ J_{n,m} = \int_0^\pi \frac{x^m(1-\cos nx)}{1- \cos x} dx $$

For instance, i noticed that the first integral is convergent for any value of $n$ since $\lim_{x\to0} \frac{1- \cos nx}{1 - \cos x}= n^2$. This fact may also be extended to the second integral, as well.

share|improve this question
2  
If you can solve $\int_0^\pi x^m \cos^k x$ for all $k$, you can use Chebyshev polynomals, $T_n$, defined so that $T_n(\cos x) = \cos nx$, and then compute the polynomial $\frac{T_n(y)-1}{y-1}$. –  Thomas Andrews Jul 20 '12 at 20:03
    
You can compute $I_n$ quickly with the residue theorem –  Cocopuffs Jul 20 '12 at 20:23
    
@ Cocopuffs: i'd like so much to solve them elementarily by only using high school knowledge (if that is possible). Residue theorem could be a way you may post it if you're willing. –  Chris's sis Jul 20 '12 at 20:24
    
I'm with @Thomas; judicious use of Chebyshev will likely crack this with minimum fuss. –  J. M. Jul 20 '12 at 22:47
    
I think the first integral may be elegantly solved by using $I_{n}=\frac{I_{n-1}+I_{n+1}}{2}$ –  Chris's sis Jul 24 '12 at 20:51

4 Answers 4

up vote 5 down vote accepted

Here's a way to find $I_n$ with the residue theorem: first, by symmetry, $I_n = \frac{1}{2} \int_0^{2\pi} \frac{1 - \cos(nx)}{1- \cos(x)}dx$. Consider $z = e^{ix}$ such that the integral with now be on the edge of the unit circle $C$, and use the fact that $$\cos(nx) = \frac{e^{inx} + e^{-inx}}{2}.$$ as well as $\frac{dz}{dx} = i e^{ix}$. So $dx = \frac{1}{iz} dz$ and

$$\int_0^{2\pi} \frac{1 - \cos(nx)}{1 - \cos(x)} dx = \int_C \frac{1}{iz} \frac{1 - \frac{1}{2}(z^{n} + z^{-n})}{1 - \frac{1}{2}(z + z^{-1})}dz$$ $$= \frac{1}{i}\int_C \frac{1}{z^{n}} \left(\frac{z^n - 1}{z - 1}\right)^2 dz$$ $$= \frac{2\pi i}{i} \mathrm{Res}_{z = 0} \frac{1}{z^{n}} (1 + ... + z^{n-1})^2 = 2\pi n$$

and so $I_n = \pi n$.

share|improve this answer

As for the first integral $$ I_n=\int\limits_{0}^\pi\frac{1-\cos nx}{1-\cos x}dx= \int\limits_{0}^\pi\frac{\sin^2 (nx/2)}{\sin^2(x/2)}dx= 2\int\limits_{0}^{\pi/2}\frac{\sin^2 (nx)}{\sin^2(x)}dx= 2\int\limits_{0}^{\pi/2}\left(\frac{\sin (nx)}{\sin(x)}\right)^2dx= $$ $$ 2\int\limits_{0}^{\pi/2}\left(\frac{e^{inx}-e^{-inx}}{e^{ix}-e^{-ix}}\right)^2dx= 2\int\limits_{0}^{\pi/2}\left(\sum\limits_{k=0}^{n-1}(e^{ix})^k (e^{-ix})^{n-k-1}\right)^2dx= $$ $$ 2\int\limits_{0}^{\pi/2}\left(\sum\limits_{k=0}^{n-1}(e^{ix})^{2k+1-n} \right)^2dx= 2\int\limits_{0}^{\pi/2}\left(\sum\limits_{k=0}^{n-1}(e^{ix})^{2k+1-n} \right)\left(\sum\limits_{l=0}^{n-1}(e^{ix})^{2l+1-n} \right)dx= $$ $$ 2\int\limits_{0}^{\pi/2}\sum\limits_{k=0}^{n-1}\sum\limits_{l=0}^{n-1}(e^{ix})^{2k+1-n+2l+1-n}dx= 2\sum\limits_{k=0}^{n-1}\sum\limits_{l=0}^{n-1}\int\limits_{0}^{\pi/2}(e^{ix})^{2k+2l+2-2n}dx= $$ $$ 2\sum\limits_{k=0}^{n-1}\sum\limits_{l=0}^{n-1}\frac{\pi}{2}\delta_{2k+2l+2-2n}= \pi\sum\limits_{k=0}^{n-1}\sum\limits_{l=0}^{n-1}\delta_{2k+2l+2-2n}=\pi n $$ As for the second... I'll think about it.

share|improve this answer
    
nice your way! (+1) –  Chris's sis Jul 20 '12 at 20:27

The exponential generating function of $J_{m,n}$ for fixed $n$ is $$G_n(t) = \sum_{m=0}^\infty J_{m,n} t^m/m! = \int_0^\pi e^{tx} \dfrac{\sin(nx)^2}{\sin(x)^2}\ dx$$ We then have $(\sin(nx)/\sin(x))^2 = n + 2 \sum_{j=1}^{n-1} (n-j) \cos(2 j x)$ and $$G_n(t) = (e^{\pi t} - 1) \left(\frac{n}{t} + 2 \sum_{j=1}^{n-1} (n-j) \dfrac{t}{t^2 + 4 j^2}\right)$$

EDIT: For example, $G_3(t) = (e^{\pi t} - 1) \left(\dfrac{3}{t} + \dfrac{4t}{t^2+4} + \dfrac{2t}{t^2+16}\right)$. Now $$e^{\pi t} - 1 = \sum_{j=1}^\infty \pi^j \dfrac{t^j}{j!} = \pi t + \dfrac{\pi^2 t^2}{2} + \dfrac{\pi^3 t^3}{6} + \ldots$$ $$\dfrac{3}{t} + \dfrac{4t}{t^2+4} + \dfrac{2t}{t^2+16} = \dfrac{3}{t} + \sum_{j=0}^\infty (-1)^j(4^{-j}+ 16^{-j}/8) t^{2j+1} = \dfrac{3}{t} + \left(1+\dfrac{1}{8}\right) t - \left(\dfrac{1}{4} + \dfrac{1}{16 \times 8}\right) t^3 + \ldots$$ and so $J_{m,3}$ is $m!$ times the coefficient of $t^m$ in the product of these.

share|improve this answer
    
So it is remains to take a limits $t\to+0$? –  Norbert Jul 20 '12 at 21:15
    
@Norbert: no, the limit as $t \to 0$ is just $G_n(0) = J_{0,n}$. –  Robert Israel Jul 20 '12 at 21:17
    
Agree, that was my stupid comment –  Norbert Jul 20 '12 at 21:17

Another generating function approach is to set $z=e^{ix}$ in $$ \begin{align} &\sum_{k=0}^\infty\frac{(it)^m}{m!}\int_0^{\pi}x^m\frac{1-\cos(nx)}{1-\cos(x)}\,\mathrm{d}x\\ &=\int_0^{\pi}z^t\left(\frac{z^n-1}{z-1}\right)^2\frac{\mathrm{d}z}{iz^n}\\ &=\frac1i\int_0^{\pi}\left(\sum_{k=1-n}^{n-1}(n-|k|)z^{k+t-1}\right)\mathrm{d}z\\ &=\frac1i\sum_{k=1-n}^{n-1}(n-|k|)\frac{(-1)^ke^{\pi it}-1}{k+t}\\ &=in\frac{1-e^{\pi it}}{t}-2it\sum_{k=1}^{n-1}\frac{n-k}{k^2-t^2}\left(1-(-1)^ke^{\pi it}\right)\tag{1} \end{align} $$ As $t\to0$, it is easy to see that we get $\pi n$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.