Improper Integral: $\int_{-\infty}^\infty\frac{e^{-t}}{1+e^{-2t}}\ dt$

Question

$$\int_{-\infty}^\infty\frac{e^{-t}}{1+e^{-2t}}\ dt$$

I have the antiderivative as

$$-\arctan e^{-t}$$

but when I do it out, I end up getting $$-\frac\pi4 + 0 - \frac\pi2+\frac\pi4$$

However, I understand that the answer should be $$\frac{\pi}2$$

Did I get the antiderivative wrong or something?

Answer 1

Nope. Just a sign error. $$\int_0^\infty+\int_{-\infty}^0=0-(-\pi/4)+(-(-\pi/4)-0)=\pi/2.$$

Answer 2

$$\begin{align} \int_{-\infty}^{\infty}\frac{e^{-t}}{1+e^{-2t}}dt&=\lim_{L\to \infty}\left(\left .-\arctan (e^{-t})\right|_{-L}^{L}\right)\\\\ &=-\arctan(0)+\lim_{L\to \infty}\arctan(e^{L})\\\\ &=\frac{\pi}{2} \end{align}$$

Answer 3

$$\left.-\tan^{-1} e^{-t}\right|^\infty_{-\infty}=(-\tan^{-1}0)-(-\tan^{-1}\infty)=0-(-\frac\pi 2)=\frac\pi 2$$