What is the value this integral ?

Thanks to Euler, we know that

\[ \begin{aligned} \sin{\theta} &= \frac{e^{i\theta}-e^{-i\theta}}{2i} \\ \cos{\theta} &= \frac{e^{i\theta}+e^{-i\theta}}{2} \\ \therefore \sin{nx} &= \frac{e^{inx}-e^{-inx}}{2i} \\ \cos{mx} &= \frac{e^{imx}+e^{-imx}}{2} \\ \end{aligned} \]

Therefore

\[ \begin{aligned} \int_{-\pi}^{\pi}\sin{nx}\cos{mx}\,dx &= \int_{-\pi}^{\pi} \frac{e^{inx}-e^{-inx}}{2i} \cdot \frac{e^{imx}+e^{-imx}}{2} \,dx \\ &= \frac{1}{4i} \int_{-\pi}^{\pi} \left( e^{i(n+m)x} + e^{i(n-m)x} - e^{-i(n-m)x} - e^{-i(n+m)x} \right)\,dx \\ &= \left[-\frac{e^{i(n+m)x}}{4(n+m)} - \frac{e^{i(n-m)x}}{4(n-m)} - \frac{e^{-i(n-m)x}}{4(n-m)} - \frac{e^{-i(n+m)x}}{4(n+m)} \right]_{-\pi}^{\pi} \\ &= -\frac{1}{4(n+m)}\left[ e^{i(n+m)x} + e^{-i(n+m)x} \right]_{-\pi}^{\pi} -\frac{1}{4(n-m)}\left[ e^{i(n-m)x} + e^{-i(n-m)x} \right]_{-\pi}^{\pi} \\ \end{aligned} \]

But anything of the form:

\[ \begin{aligned} \left[e^{i\beta x} + e^{-i\beta x}\right]_{-\pi}^{\pi} &= (\cos{\beta\pi} + i\sin{\beta\pi}) + (\cos{(-\beta\pi)} + i\sin{(-\beta\pi)}) - \left[ (\cos{(-\beta\pi)} + i\sin{(-\beta\pi)}) + (\cos{\beta\pi} + i\sin{\beta\pi}) \right] \\ &= \cos{\beta\pi} + i\sin{\beta\pi} + \cos{\beta\pi} - i\sin{\beta\pi} - \left[ \cos{\beta\pi} - i\sin{\beta\pi} + \cos{\beta\pi} + i\sin{\beta\pi} \right] \\ &= 0 & ! \end{aligned} \]

Therefore,

\[ \begin{aligned} \int_{-\pi}^{\pi}\sin{nx}\cos{mx}\,dx &= 0 \\ \end{aligned} \]

\(\Box\)

Who would have thought ?

# posted by rot13(Unafba Pune) @ 7:23 AM