Integrating a function around three branches using a semi-circle contour

[Ex] Watson’s complex integration

$$\int^{\pi/2}_{0}\cos(nt)\cos^m(t)\,dt=\frac{\pi \Gamma(m+1)}{2^{m+1}\Gamma\left(\frac{n+m+2}{2}\right)\Gamma\left(\frac{2-n+m}{2}\right)}$$

$$\textit {solution}$$

Let us integrate the following function

$$f(z) = z^{n-m-1}\left(1+z^2\right)^m$$

We choose the principle logarithm where

$$\log(z) = \log|z|+\mathrm{Arg}(z)$$

Note that the function \(z^{n-m-1} = e^{(n-m-1)\log(z)}\) will have a branch cut on the negative \(x\) axis. Also we have

$$\left(1+z^2 \right)^m = e^{m\log(1+z^2)}$$

The principle branch will be on the imaginary \(y\) axis where \( |y| \geq 1\) which implies that \( y \geq 1\) or \(y \leq -1\)

By the residue theorem

$$\int_{\gamma_{\epsilon_1}}f(z)\,dz+\int_{\gamma_{\epsilon_2}}f(z)\,dz+\int_{\gamma_{\epsilon_3}}f(z)\,dz\\+\int_{i-i\epsilon_1}^{i\epsilon_2}f(z)\,dz+\int_{-i\epsilon_2}^{-i+i\epsilon_3}f(z)\,dz+\int_{C}f(z)\,dz=0$$
I’ll prove the second integral goes to 0

Use the parametrization \(\gamma_{\epsilon_2}(t)= \epsilon_2e^{it}\,\,,-\pi/2\leq t \leq \,\pi/2\)
$$\left|i\epsilon_2^{n-m}\int_{-\pi/2}^{\pi/2}e^{int}\log(1+(\epsilon_2e^{it})^2)\,dt\right|\leq \pi \log(2)\epsilon_2^{n-m} $$

By taking the limit we deduce the integral goes to 0. Similarly we have

$$\lim_{\epsilon_1 \to 0}\int_{\gamma_{\epsilon_1}}f(z)\,dz=\lim_{\epsilon_2 \to 0}\int_{\gamma_{\epsilon_3}}f(z)\,dz=0$$

Note on on the circular part of \(|z|=1\)

\begin{align}\int_{C}f(z)\,dz &=i\,\int^{\pi/2}_{-\pi/2}e^{it}e^{int-imt-it}\left(1+e^{2it}\right)^m\,dt \\ &=i\,\int^{\pi/2}_{-\pi/2}e^{int}(e^{-it}+e^{it})^m\,dt\\ &=i2^m\, \int^{\pi/2}_{-\pi/2}e^{int}\cos^m(t)\,dt \end{align}

The integrals reduce to

$$\int_{i}^{0}x^{n-m-1}\left(1+x^2 \right)^m\,dx+\int_{0}^{-i}x^{n-m-1}\left(1+x^2 \right)^m\,dx=-i2^m\int^{\pi/2}_{-\pi/2}e^{int}\cos^m(t)\,dt$$

By taking \(x=it\) and \(x=-it\) respectively

$$-i\int_{0}^{1}(it)^{n-m-1}\left(1-t^2 \right)^m\,dt-i\int_{0}^{1}(-it)^{n-m-1}\left(1-t^2 \right)^m\,dt=-i2^m\int^{\pi/2}_{-\pi/2}e^{int}\cos^m(t)\,dt$$

We then can combine the integrals
$$-i((i)^{n-m-1}+ (-i)^{n-m-1})\int_{0}^{1}t^{n-m-1}\left(1-t^2 \right)^m\,dt=-i2^m\int^{\pi/2}_{-\pi/2}e^{int}\cos^m(t)\,dt$$

Note that

$$-(i^{n-m}-(-i)^{n-m})= -\left(e^{i(n-m)\pi/2} -e^{-i(n-m)\pi/2} \right)=-2i\sin\left(\frac{n\pi -m\pi}{2} \right)$$

We deduce that

$$\int^{\pi/2}_{-\pi/2}e^{int}\cos^m(t)\,dt=2^{1-m}\sin\left(\frac{n\pi -m\pi}{2} \right)\int^1_{0}t^{n-m-1}(1-t^2)^m\,dt$$

Using the Euler formula we have

$$\int^{\pi/2}_{0}\cos(nt)\cos^m(t)\,dt=2^{-m}\sin\left(\frac{n\pi -m\pi}{2} \right)\int^1_{0}t^{n-m-1}(1-t^2)^m\,dt$$

Note the beta integral

$$\int^1_{0}t^{n-m-1}(1-t^2)^m\,dt = \frac{\Gamma(m+1)\Gamma\left( \frac{n-m}{2}\right)}{2\Gamma\left(\frac{n+m+2}{2}\right)}$$

By the reflection formula we have

$$\Gamma\left( \frac{n-m}{2}\right)\Gamma\left(1- \frac{n-m}{2}\right)=\frac{\pi}{\sin\left(\frac{n\pi -m\pi}{2} \right)}$$

We deduce then that

$$\int^{\pi/2}_{0}\cos(nt)\cos^m(t)\,dt=\frac{\pi \Gamma(m+1)}{2^{m+1}\Gamma\left(\frac{n+m+2}{2}\right)\Gamma\left(\frac{2-n+m}{2}\right)}$$

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