Integrate sin(nx)sin(mx) from 0 to 2$\pi$ using residuals

Question

I need to use the residues integration method to calculate the following integral:

$\int_0^{2\pi} \sin(nx)\sin(mx)dx$ where m and n are positive whole numbers.

I know that I need to transform the sinus into its exponential form then substitute for $z=e^{ix}$ and do a change of variable in the integral, then use the theorem saying that the integral along a closed curve is $2\pi i$ times the sum of the residues of all the singularities inside the curve.

However, I do not manage to get the right answer, and for the case m=n I get $\pi/2$ instead of $\pi$.

Answer 1

\begin{align*} \int_{0}^{2\pi}\sin(nx)\sin(mx)dx&=-\dfrac{1}{4}\int_{0}^{2\pi}(e^{inx}-e^{-inx})(e^{imx}-e^{-imx})dx\\ &=-\dfrac{1}{4}\int_{0}^{2\pi}(e^{i(m+n)x}-e^{-i(m+n)x}-e^{-i(n-m)x}-e^{-i(m-n)x})dx\\ &=-\dfrac{1}{4}\int_{0}^{2\pi}(-e^{-i(n-m)x}-e^{-i(m-n)x})dx\\ &=-\dfrac{1}{4}\delta_{m,n}(-2\pi-2\pi)\\ &=\delta_{m,n}\pi. \end{align*}

Answer 2

The trick for trigonometric integrals: $$ z = e^{ix}\implies\sin(nx) = \frac12(e^{inx} − e^{−inx}) = \frac12(z^n - z^{-n}) \qquad dz = iz\,dx $$ $$ \int_{0}^{2\pi}\sin(nx)\sin(mx)\,dx = \frac14\int_{|z|=1}(z^n - z^{-n})(z^m - z^{-m})\frac1{iz}\,dz = $$ $$ \frac1{4i}\int_{|z|=1}(z^{m-n-1} + z^{n-m-1} - z^{m+n-1} - z^{-m-n-1})\,dz = \cdots $$ In your problematic case: $$ m = n\ne 0\implies z^{m-n-1} = z^{n-m-1} = z^{-1}, z^{m+n-1}\ne z^{-1}, z^{-m-n-1}\ne z^{-1}, $$ and the integral is $\frac{2\cdot2\pi i}{4i} = \pi$.

Answer 3

Hints:

• If $f(a-x)=f(x)$ on $[0,a]$, then $\int_0^a f(x)~\mathrm dx=\int_0^a f(a-x)~\mathrm dx$

• If $f(2a-x)=f(x)$ on $[0,2a]$, then $\int_0^{2a} f(x)~\mathrm dx=2\int_0^a f(x)~\mathrm dx$

• Suppose $m=n=k=2^rs$ where $2\not\mid s$. Can you show that $$\int_0^{2\pi}\sin^2(kx)~\mathrm dx=2^{r+1}\int\limits_0^{\pi/2^r}\sin^2(kx)~\mathrm dx=2^{r+1}\cdot\frac{\dfrac\pi{2^r}-0}2=2^{r+1}\cdot\frac\pi{2^{r+1}}=\pi$$