Understanding an equality involving expoentials and integrals

Question

Why is it that if $n \in \mathbb{Z}, g_n(x)=e^{in}$, then $\int g_n(x)e^{-imx}dx= 0$ if $n=m$ or $1$ if $n≠m$

Now let $n=m$, then $\int g_n(x)e^{-imx}dx = \int e^{inx}e^{-imx}dx=\int e^0dx= \int dx$

Now let $n≠m$ then $\int g_n(x)e^{-imx}dx = \int e^{inx}e^{-imx}dx=\int e^{ix(n-m)}dx$

But I really can't go farther.

Could someone explain why $\int g_n(x)e^{-imx}dx= 0$ if $n=m$ or $1$ if $n≠m$

Answer 1

hint

$$e^{ipx}=\cos (px)+i\sin (px) $$

$$\int_0^{2\pi}\cos (px)dx=0$$

Answer 2

What you are missing are the limits - usually it is written as $$\int_{-\pi}^\pi e^{inx}e^{-imx}\,dx=\int_{-\pi}^\pi e^{i(n-m)x\,dx}=\begin{cases}\int_{-\pi}^\pi 1\,dx&\text{if } n=m\\\int_{-\pi}^\pi e^{ikx}\,dx&\text{if }n\neq m\end{cases}=\begin{cases}2\pi&n=m\\\frac{1}{ik}(e^{ik\pi}-e^{-ik\pi})&n\neq m \end{cases}$$

This last step gives $0$ for the case $n\neq m $ since $k\in\Bbb Z$, so $e^{ik\pi}=(-1)^k=(-1)^{-k}=e^{-ik\pi}$.

So the actual result is $$\int_{-\pi}^\pi e^{inx}e^{-imx}\,dx=\begin{cases}2\pi&n=m\\0&n\neq m \end{cases}=2\pi\delta_{nm}$$

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Alternatively you can also use limits $0\to2\pi$.