In comparing improper integrals, what is a similar integrand to ∫ (1/(e^(2x)+1))dx from a = 1 to b = ∞. I know it converges, I just can't find a similar integrand.

Question

amistre64 · Answer

$$\lim_{b->\infty}\ \int_{1}^{b}\frac{1}{e^{2x}+1}dx$$
right?

zarkon · Answer

if you are just trying to prove convergence ....compare it to $$\frac{1}{e^{2x}}$$

myininaya · Answer

can we just integrate lol

anonymous · Answer

Zarkon, that's what I had at first, just input it wrong.. Whoops.
Thanks.

myininaya · Answer

$$\int\limits_{}^{}\frac{1}{e^{2x}+1} \cdot \frac{e^{-2x}}{e^{-2x}} dx=\int\limits_{}^{}\frac{e^{-2x}}{1+e^{-2x}} dx$$

let $$u=1+e^{-2x}=> du=-2 e^{-2x} dx$$

$$\int\limits_{}^{}\frac{e^{-2x}}{1+e^{-2x}} dx=\frac{-1}{2}\int\limits_{}^{}\frac{du}{u}=\frac{-1}{2}\ln|u|+C$$
$$=\frac{-1}{2}\ln(1+e^{-2x})+C$$

myininaya · Answer

whats the point in comparing it when it is easy to integrate

myininaya · Answer

$$\lim_{b \rightarrow \infty}\int\limits_{1}^{b}\frac{1}{e^{2x}+1} dx=\lim_{b \rightarrow \infty}[\frac{-1}{2}\ln(1+e^{-2x})]_1^b$$