Quick fun definite integral

In the interval $(0,\infty)$ the function $\mathrm{e}^{-x} <1$ . Keeping this in mind consider the expression for all $\lvert y \rvert<1$ :

$\ln\left(1+y\right)=y - \frac{y^2}{2} +\frac{y^3}{3} + \dots$

Since $\mathrm{e}^{-x}<1$ in the given interval, replacing $y=\mathrm{e}^{-x}$ gives:

$\ln\left(1+\mathrm{e}^{-x}\right)=\mathrm{e}^{-x} - \frac{\mathrm{e}^{-2x}}{2} +\frac{\mathrm{e}^{-3x}}{3} + \dots$

Integrating both sides with respect to $x$ as such:

$\int_{0}^{\infty} \ln\left(1+\mathrm{e}^{-x}\right) \ dx = \int_{0}^{\infty} \left(\mathrm{e}^{-x} - \frac{\mathrm{e}^{-2x}}{2} +\frac{\mathrm{e}^{-3x}}{3} + \dots\right) \ dx$

Denoting the above integral as $S$ :

$\implies S = \int_{0}^{\infty} \ln\left(1+\mathrm{e}^{-x}\right) \ dx = 1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \dots$

We are familiar with the following sum:

$S_1=\frac{\pi^2}{6} = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \dots$

$\implies S_1 = \left(1 + \frac{1}{3^2} + \dots\right)+\left(\frac{1}{2^2} + \frac{1}{4^2} + \dots\right)$

$\implies S_1 = \left(1 + \frac{1}{3^2} + \dots\right)+\frac{1}{2^2}\left(1+\frac{1}{2^2} + \frac{1}{3^2} + \dots\right)$

$\implies S_1 = \left(1 + \frac{1}{3^2} + \frac{1}{5^2}+\dots\right)+\frac{S_1}{4}$

$\implies \frac{3S_1}{4} = 1 + \frac{1}{3^2} + \frac{1}{5^2}+\dots$

Now consider $S$ :

$S = 1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \dots$ $\implies S = \left(1 + \frac{1}{3^2} + \dots\right) - \left(\frac{1}{2^2} + \frac{1}{4^2} + \dots\right)$ $\implies S = \frac{3S_1}{4} - \frac{1}{4}S_1$

$\implies S = \frac{S_1}{2}$ $\implies S = \frac{\pi^2}{12}$

$\therefore \int_{0}^{\infty} \ln\left(1+\mathrm{e}^{-x}\right) \ dx= \frac{\pi^2}{12}$

Similar solution as @Karan Chatrath 's

$\begin{aligned} I & = \int_0^\infty \ln(1+e^{-x}) \ dx & \small \blue{\text{By Maclaurin series}} \\ & = \int_0^\infty \left(e^{-x} - \frac {e^{-2x}}2 + \frac {e^{-3x}}3 - \frac {e^{-4x}}4 + \cdots \right) dx \\ & = - e^{-x} + \frac {e^{-2x}}{2^2} - \frac {e^{-3x}}{3^2} + \frac {e^{-4x}}{4^2} - \cdots \bigg|_0^\infty \\ & = \frac 1{1^2} - \frac 1{2^2} + \frac 1{3^2} - \frac 1{4^2} + \cdots \\ & = \left(\frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \cdots \right) - 2 \left(\frac 1{2^2} + \frac 1{4^2} + \frac 1{6^2} + \cdots \right) \\ & = \left(\frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \cdots \right) - \frac 2{2^2} \left(\frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \cdots \right) \\ & = \frac 12 \left(\frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \frac 1{4^2} \cdots \right) & \small \blue{\text{Riemann zeta function }\zeta (s) = \sum_{k=1}^\infty \frac 1{k^s}} \\ & = \frac {\zeta (2)}2 = \frac {\pi^2}{12} & \small \blue{\text{and }\zeta(2) = \frac {\pi^2} 6} \end{aligned}$

Therefore, $a+b = 2+12=\boxed{14}$ .

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References:

• Maclaurin series
• Riemann zeta function