Show $\int_0^{2\pi}\cos(n\phi')\cos^l(\phi-\phi')\mathrm{d}\phi=\frac{2\pi}{2^l}\cos(l\phi)\delta_{l,n}$

Question

I have to show $$\int_0^{2\pi}\cos(n\phi')\cos^l(\phi-\phi')\mathrm{d}\phi=\frac{2\pi}{2^l}\cos(l\phi)\delta_{l,n}$$ where $l,n$ are positive integers such that $l\leq n$ I'm supposed to use the fact

$$ \int_0^{2\pi}e^{i(l-n)\phi}\mathrm{d}\phi=2\pi\delta_{l,n}$$

But I'm really lost. I tried to rewrite the cosines as the real part of $e^{in\phi}$ et cetera and trying to expand the power with the binomial theorem but it didn't work. I don't see any other way to obtain a useful way to use that identity. Any input will be appreciated

Functions plural? I know $\Gamma(x)=\int_0^\infty t^{x-1}e^{-t}\mathrm{d}t$ but I don't know if you mean this Gamma function — user438666, Nov 05 '17 at 19:03
I don't see how it could be useful. Could you elaborate please? — user438666, Nov 05 '17 at 19:24

score 1 · Accepted Answer · answered Nov 05 '17 at 19:25

1

Use $\cos x=(e^{ix}+e^{-ix})/2$ to rewrite $$ \int_0^{2\pi}\cos(n\phi')\cos^l(\phi-\phi')\mathrm{d}\phi'=\frac{1}{2^{l+1}}\int_0^{2\pi}(e^{in\phi'}+e^{-in\phi'})(e^{i(\phi-\phi')}+e^{-i(\phi-\phi')})^l d\phi'\ . $$ Then use the binomial theorem to rewrite $$ \frac{1}{2^{l+1}}\sum_{k=0}^l {l\choose k}\int_0^{2\pi}(e^{in\phi'}+e^{-in\phi'})e^{ik(\phi-\phi')}e^{-i(l-k)(\phi-\phi')} d\phi'=\frac{1}{2^{l+1}}\sum_{k=0}^l {l\choose k} e^{i\phi(2k-l)}2\pi(\delta_{n-k+(l-k),0}+\delta_{-n-k+(l-k),0})=\frac{2\pi}{2^{l+1}}\left[{l\choose (l+n)/2}e^{in\phi}+{l\choose (l-n)/2}e^{-in\phi}\right]\ , $$ from which you should be able to complete.

answered Nov 05 '17 at 19:25

Pierpaolo Vivo

6,376

I'm confused by your last line, we never assumed that $l+n$ or $l-n$ are even – user438666 Nov 05 '17 at 19:42
Me neither, actually... – Pierpaolo Vivo Nov 05 '17 at 19:45
Shouldn't the coefficients in the binomial coefficient be integers? – user438666 Nov 05 '17 at 19:46
1

Not necessarily https://math.stackexchange.com/questions/340124/binomial-coefficients-1-2-k – Pierpaolo Vivo Nov 05 '17 at 19:47
I can see that for $l=n$ we obtain the desired result. I cannot see why it's 0 otherwise – user438666 Nov 05 '17 at 19:51
Got it in the end. The terms are not in the sum if $b\neq n$ because $(b+n)/2\geq b$ and $(b-n)/2\leq 0$ because $b\leq n$. Thank you. – user438666 Nov 05 '17 at 20:31

Show $\int_0^{2\pi}\cos(n\phi')\cos^l(\phi-\phi')\mathrm{d}\phi=\frac{2\pi}{2^l}\cos(l\phi)\delta_{l,n}$

1 Answers1