TITLE: prove integration of $\cos nx \cos mx / \sqrt (1 − x^2)$ from $-1$ to $1$ is $0$
QUESTION [0 upvotes]: I am struggling with this prove integral is $0$ question, I would really appreciate anyone that could help.
The questions asks to prove that for m ≠ n：
$$\int_{-1}^1\frac{\cos nx\cos mx}{\sqrt{1 − x^2}}\  dx = 0$$
enter image description here = 0
I know the integral is twice the integral from $0$ to $1$, but I have no idea how to prove it.because cos is not an odd function so the integral over a symmetric interval is not $0$...

REPLY [1 votes]: It's no wonder you are struggling:  What you are trying to prove is simply not true in general.
The quickest way to see this is to consider the special case $n=0$, $m=1$.  The integrand $\cos x\over\sqrt{1-x^2}$ is strictly positive for $-1\lt x\lt1$, hence
$$\int_{-1}^1{\cos x\over\sqrt{1-x^2}}dx\gt0$$
If you dismiss this counterexample by requiring $n$ and $m$ to both be positive, then consider $n=1$ and $m=2$.  Wolfram Alpha finds
$$\int_{-1}^1{\cos x\cos2x\over\sqrt{1-x^2}}dx\approx0.793481$$
(and an exact value in terms of Bessel functions).  It shouldn't be too difficult to give a non-Wolfram proof that this integral is positive, but I don't see any easy way to make it obvious.
It seems likely that the integral in question is never equal to $0$.  Wolfram Alpha finds in general that
$$\int_{-1}^1{\cos nx\cos mx\over\sqrt{1-x^2}}dx={1\over2}\pi(J_0(m-n)+J_0(m+n))$$
where $J_0$ is a Bessel function of the first kind.  A hand-wavy reason why you don't get $0$ out of this is that, while the limits of integration here, $\pm1$, are "natural" values for the square root function in the denominator, they are not particularly natural for the cosine functions in the numerator, for which rational multiples of $\pi$ are more natural.  There might be an actual proof based on properties of Bessel functions; I'd love to see one.
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