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I have encountered such a integral formula that I cannot prove. $$\int_{-\pi}^{\pi}\cos(m(x-y))\cos(n(x-y))dx=\pi\delta_{mn}(2-H_1(m))$$ where $H_1(m)=\begin{cases} 0 & \text{if } m=0\\ 1 & \text{if } m\geq1 \end{cases}$

If being converted to the known form $\int\cos mx\cos nx dx$, but then the bounds become very complicated \begin{align} I(t)&=\int_{-\pi}^{\pi}\cos(m(x-y))\cos(n(x-y))dx\\ &=\int_{-\pi-y}^{\pi-y}\cos(mt)\cos(nt)dt\\ &=\frac{1}{2}\int_{-\pi-y}^{\pi-y}\cos((m+n)t)+\cos((m-n)t)dt\\ &=\left|\frac{\sin(m+n)t}{2(m+n)}\right|_{-\pi-y}^{\pi-y}+\left|\frac{\sin(m-n)t}{2(m-n)}\right|_{-\pi-y}^{\pi-y} \end{align}

MathArt
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1 Answers1

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Your formula for $I(t)$ is only valid for $m \ne n$, $m+n \ne 0$. Moreover for any integer $k$:

$$\sin (k(\pi-y)) = \sin(k\pi-ky-2k\pi)= \sin(k(-\pi-y))$$

hence $I(t) = 0$ for $m\ne n$.

For $m=n\ne 0$. we arrive at

$$I(t) = \frac 12\int^{\pi-y}_{-\pi-y}(\cos ((m+n)t) + \cos 0)\ dt = \frac 12\int^{\pi-y}_{-\pi-y}(\cos (2mt) +1)\ dt$$

The value of the integral for the cosine term in front, as demonstrated, is $0$.

Hence $$I(t) = \frac12\int^{\pi-y}_{-\pi-y}1\ dt = \pi$$

For $m=n=0$, the integral is simply $\displaystyle \int^{\pi}_{-\pi}1\ dx = 2\pi$.

These three cases matches up with $\pi\delta_{mn}(2-H_1(m))$.

player3236
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