Integral $\int_0^{\frac{-\pi}{2}} \frac{(2\pi i + \epsilon e^{i\theta})^4i\epsilon e^{i\theta} d \theta} {e^{2 \pi i + \epsilon e^{i \theta}} - 1}$

Question

I'm not sure to see how the integral $$\int_0^{\frac{-\pi}{2}} \frac{(2\pi i + \epsilon e^{i\theta})^4i\epsilon e^{i\theta} d \theta} {e^{2 \pi i + \epsilon e^{i \theta}} - 1}$$ yields a finite value of $-8\pi^5i$

I was following this post to solve a contour integral. After expanding the power and neglected all $\epsilon$ terms with a power of 2 or more. I get $$ \int_0^{\frac{-\pi}{2}} \frac{16 \pi^4 i \epsilon e^{i \theta} d \theta}{e^{2 \pi i + \epsilon e^{i \theta}} - 1}$$

From here I'm not sure if I can apply any other simplification.

Take $\epsilon \to 0^+$ because the quarter-circle arc shrinks to recover the original integral from that linked post. You'd see an indeterminate form $\frac{0}{0}$ which prompts L'Hôpital's rule. — Accelerator, Jun 22 '24 at 19:52
The denominator become $e^{2 \pi i} - 1$ which is not 0. I don't see the indeterminate form. — epselonzero, Jun 22 '24 at 21:08

score 1 · Answer 1 · answered Jun 23 '24 at 06:36

Using $e^{2\pi i}=1$, let $\theta=-i\log(x)$ and then $x \epsilon=u$, to face $$I=\int\frac{(u+2 i \pi )^4}{e^u-1}\,du$$ which will give a series of polylogarithms.

Back to $\theta$ and the bounds, use Taylor series for small values of $\epsilon$ to obtain for the definite integral $$J=-8 i \pi ^5+8(1+ i) \pi ^3 (\pi +4 i)\, \epsilon+O( \epsilon)^2$$ that you can simplify.

Integral $\int_0^{\frac{-\pi}{2}} \frac{(2\pi i + \epsilon e^{i\theta})^4i\epsilon e^{i\theta} d \theta} {e^{2 \pi i + \epsilon e^{i \theta}} - 1}$

1 Answers1