Find $\int_0^{\infty} \frac{dx}{1+e^x}$

Question

$$\int_1^\infty\frac{dx}{1+e^x} $$

$$\lim_{M\to\infty}\int_1^M\frac{e^xdx}{e^x(1+e^x)} \\ u= 1 + e^x \\ du = e^x dx \\ \lim_{M\to\infty} \int_{1+e}^{1+e^M} \frac{du}{(u-1)u} $$ I then found the partial fractions which gave me $$\lim_{M\to\infty} \int_{1+e}^{1+e^M} \frac{du}{(u-1)} - \int_{1+e}^{1+e^M} \frac{du}{u}$$

$$\lim_{M\to\infty} ln(e^M) - lne - ln(e^M + 1) + ln(1 + e)$$ That's where I'm a bit confused because I can get either $$1.\lim_{M\to\infty} ln(e^M) - 1 + ln(\frac{1 + e}{e^M + 1}) $$ which gives me$$ \infty - 1 + \infty = \infty$$

or $$2.\lim_{M\to\infty} ln(\frac{e^M}{e^M + 1}) + ln(\frac{1 + e}{e}) $$ which I thought would equal to $$\infty$$ since $$\lim_{M\to\infty} ln(\frac{e^M}{e^M + 1}) + ln(\frac{1 + e}{e}) = \infty + ln(\frac{1 + e}{e}) = \infty$$ But my professor said that the answer is just $$ln(\frac{1 + e}{e})$$ I did not understand his explanation. Any clarification would be great.

Answer 1

We have

$$\lim_{M \to \infty} \ln \left( \frac{e^M}{e^ M + 1} \right) = \lim_{M \to \infty} \ln \left( \frac{1}{1 + e^{-M}} \right) = \log \left( \lim_{M\to\infty} \frac{1}{1 + e^{-M}}\right)$$

The last equality holds because the limit $\lim_{M\to\infty} \frac{1}{1 + e^{-M}}$ exists and is in the domain of $\log$, which is continuous. Thus to finish

$$\log \left( \lim_{M\to\infty} \frac{1}{1 + e^{-M}}\right) = \log 1 = 0.$$

Answer 2

$\lim_{M\to \infty}ln(e^M)−lne−ln(e^M+1)+ln(1+e)$ is correct. The above is equal to $ln((1+e)/e) - \lim_{M\to\infty}ln(1+1/e^M) = ln((1+e)/e)$.

In your calculations first method, you are doing $\infty - \infty$ which is not correct way to evaluate the limits. In second method, your limit is incorrect - it has to be $0$ instead of $\infty$