2
$\begingroup$

How to show (possibly without using any reduction formulae techniques) that

$$ \int_{-\pi}^\pi\cos^n(a-x)\cos((n-2r)x) \,\mathrm dx = \lambda\cos((n-2r)a) $$ with $-\pi<a<\pi$ and $n$ a positive integer, $r<\frac{n}{2}$ a positive integer.

$\endgroup$
  • 1
    $\begingroup$ What have you attempted thus far? Oh, and by the way, use \mathrm dx to generate $\mathrm dx$ instead of $dx$ if you want. $\endgroup$ – Mr Pie Feb 14 '18 at 19:42
  • $\begingroup$ @user477343 I've tried to use $\cos^n(x)=\frac{(e^{ix}+e^{-ix})^n}{2^n}$ and then Binomial theorem, but it seems to be too tedious. $\endgroup$ – Alex Feb 14 '18 at 19:49
  • $\begingroup$ Do you have any restriction on $\lambda$? If not, then $a$ could be anything, just choose $\lambda$ appropriately after picking any $a$. $\endgroup$ – Clayton Feb 14 '18 at 19:54
  • $\begingroup$ $\lambda$ is a fixed (maybe complex) parameter, the same for any $a$. $\endgroup$ – Alex Feb 14 '18 at 19:56
3
$\begingroup$

We can transform the integral as \begin{align} I&=\int_{-\pi}^\pi\cos^n(a-x)\cos((n-2r)x) \,\mathrm dx \\ &=\int_{a-\pi}^{a+\pi}\cos^n(y)\cos((n-2r)(a-y)) \,\mathrm dy \\ &=\int_{a-\pi}^{a+\pi}\cos^n(y)\left[\cos((n-2r)a)\cos((n-2r)y)+\sin((n-2r)a)\sin((n-2r)y)\right] \,\mathrm dy\\ &=\int_{-\pi}^{\pi}\cos^n(y)\left[\cos((n-2r)a)\cos((n-2r)y)+\sin((n-2r)a)\sin((n-2r)y)\right]\,\mathrm dy\\ &=\cos((n-2r)a)\int_{-\pi}^{\pi}\cos^n(y)\cos((n-2r)y)\,\mathrm dy\\ \end{align} where we used the periodicity of the functions and their parity properties. Then we have to evaluate \begin{equation} \lambda=\int_{-\pi}^{\pi}\cos^n(y)\cos((n-2r)y)\,\mathrm dy\\ \end{equation} It can be done using the residue method by integrating on the unit circle: \begin{align} \lambda&=2^{-n-1}\int_{C}\left( z+\frac{1}{z} \right)^n\left( z^{n-2r}+\frac{1}{z^{n-2r}} \right)\frac{dz}{iz}\\ &=-i2^{-n-1}\int_{C}\left( z^2+1 \right)^n\left( z^{2n-4r}+1 \right)\frac{dz}{z^{2n-2r+1}}\\ &=2^{-n}\pi\operatorname{Res}\left[z^{-2n+2r-1}\left( z^2+1 \right)^n\left( z^{2n-4r}+1 \right);z=0\right] \end{align} The residue is the coefficient of degre $2n-2r$ of the polynom $\left( z^2+1 \right)^n\left( z^{2n-4r}+1 \right)$ which is $\binom{n}{r}+\binom {n}{n-r}=2\binom{n}{r}$. Finally, \begin{equation} \lambda=2^{1-n}\pi\binom{n}{r} \end{equation}

$\endgroup$
  • $\begingroup$ Perfect! Thanks $\endgroup$ – Alex Feb 14 '18 at 21:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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