I need help with the integral $\int_{i\pi}^{3i\pi}\frac{dx}{e^{x} + 1}$

Question

While I have tried integrating the following integral $$ I = \int_{i\pi}^{3i\pi}\frac{dx}{e^{x} + 1} $$

I am getting zero, which is mostly wrong. Here is my attempt

Integrating I: $$ \int_{i\pi}^{3i\pi}\frac{dx}{e^{x} + 1} $$

Multiplying by $e^{x}$ on both the numerator and denominator, I get $$ \int_{i\pi}^{3i\pi}\frac{e^{x}dx}{(e^{x} + 1)(e^{x})} $$

Taking $u = e^{x} +1$ & $du = e^{x}$

I get the new limits at $u(i\pi)$ and $u(3i\pi)$. Both of which give the value of u at 0.

Hence, $$ \int_{0}^{0}\frac{du}{(u-1)(u)} = 0 ? $$

I know I have made a mistake. Can anyone please show where I have made a mistake and how I can correct it. Thank you.

Does this https://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3x/439856#439856 help — lab bhattacharjee, Jul 17 '21 at 16:32
The integration contour now loops around $z=+1$, which happens to contain a singularity. Have you learned about Cauchy yet? — Trebor, Jul 17 '21 at 16:34
@Trebor: I have read about it a bit , but I don't know how to apply it in this case. — Mathnewbie10325, Jul 17 '21 at 16:36
Note that $e^x+1=0$ when $x=i(2n+1)\pi$. Moreover, $e^z+1 =(z-i(2n+1)\pi)+O(((z-i(2n+1)\pi)^2)$. Hence, this integral diverges. — Mark Viola, Jul 17 '21 at 16:38

score 1 · Accepted Answer · answered Jul 17 '21 at 18:15

1

notice that $e^{ix}$ is periodic on $[0,2\pi]$, so you can shift both bounds by the same amount and you will still get zero, it is the same as: $$\int_0^{2\pi}\sin(x)dx=\int_0^{2\pi}\cos(x)=0$$ are you sure it doesnt mean the path integral

answered Jul 17 '21 at 18:15

Henry Lee

12,215

I need help with the integral $\int_{i\pi}^{3i\pi}\frac{dx}{e^{x} + 1}$

1 Answers1