Complex integral help: $\oint_C \frac{\sin(z)}{z(z-\pi/4)} dz$

Question

I'm attempting to evaluate the following complex integral: $$\oint_C \frac{\sin(z)}{z(z-\pi/4)} dz, $$ where $C$ is a circle of radius $\pi$ centred on the origin. I have calculated the residues of this function at $z=0$ and $z=\pi/4$, and then used Cauchy's Residue Theorem to evaluate the integral. However, this method gives the result as zero, which implies that the integrand in eq.(1) is analytic, which I don't think it is.

Is my method (and/or answer) for solving this integral correct, or should I be using another method?

Thanks

Answer 1

I don't know how you computed those residues, but you got the wrong value. The residue at $0$ is indeed $0$, but the residue at $\dfrac\pi4$ is $\dfrac{\sin\left(\frac\pi4\right)}{\frac\pi4}=\dfrac{2\sqrt2}\pi$. Therefore, your integral is equal to $4\sqrt2i$.

And it is a serious error to assert that if the integral was $0$, then the integrand would be analytical.

Answer 2

Using Cauchy's integral formula, singularities at $z=0,\frac{\pi}{4}$, we have

\begin{align} \int_{\vert z \vert = 1}\frac{\sin z}{z(z-\frac{\pi}{4})}dz&=2 \pi i \bigg(\frac{\sin z}{z-\frac{\pi}{4}}\bigg\vert_{z=0}+\frac{\sin z}{z}\bigg\vert_{z=\frac{\pi}{4}}\bigg)\\ &=2 \pi i (0+\frac{2\sqrt{2}}{\pi})\\ &=4i\sqrt{2} \end{align}

as needed.