MHB Integral challenge ∫ln2(1+x^(−1))dx

[lfdahl]

Evaluate the definite integral

$$\int_{0}^{1} \ln^2(1+x^{-1}) \,dx$$

 

[MountEvariste]

Spoiler

Starting with integration by parts, we have:

$\begin{aligned}\displaystyle \mathcal{I} & = \int_{0}^{1} \ln^2(1+x^{-1}) \,dx \\& = \int_{0}^{1} (x)'\ln^2(1+x^{-1}) \,dx \\&= x \ln^2(1+x^{-1})\bigg|_{x=0}^{1}+2\int_0^{1}\frac{ \ln(1+1/x)}{1+x}\,{dx} \\& = \ln^2(2)+2\int_0^{1}\frac{ \ln(1+x)}{1+x}\,{dx}-2\int_0^{1}\frac{ \ln(x)}{1+x}\,{dx} \\& = \ln^2(2)+\ln^2(1+x)\bigg|_{x=0}^{1}-2\int_0^{1}\frac{\ln(x)}{1+x}\,{dx} \\& = 2\ln^2(2)-2 \int_0^{1}\frac{\ln(x)}{1+x}\,{dx}\end{aligned}$​

It remains to calculate the last integral. Let $x \mapsto 1-x$ which maps

$\displaystyle 2\int_0^{1}\frac{\ln(x)}{1+x}\,{dx} \mapsto \int_0^{1}\frac{\ln(1-x)}{x}\,{dx} $​

Then using the Maclaurin expansion of $\ln(1-x)$ we have

$\begin{aligned}2\int_0^{1}\frac{\ln(x)}{1+x}\,{dx} & = \int_0^{1}\frac{\ln(1-x)}{x}\,{dx} \\& = -\int_0^{1} \sum_{k \ge 0}\frac{x^{k}}{k+1} \,{dx} \\& = -\sum_{k \ge 0}\int_0^{1} \frac{x^{k}}{k+1}\,{dx} \\& = -\sum_{k \ge 0}\frac{1}{(k+1)^2} \\& = - \frac{\pi^2}{6}\end{aligned}$​

Therefore we have $\displaystyle \mathcal{I} = 2\ln^2(2)+\frac{\pi^2}{6}. $

 

[lfdahl]

Spoiler

Starting with integration by parts, we have:

$\begin{aligned}\displaystyle \mathcal{I} & = \int_{0}^{1} \ln^2(1+x^{-1}) \,dx \\& = \int_{0}^{1} (x)'\ln^2(1+x^{-1}) \,dx \\&= x \ln^2(1+x^{-1})\bigg|_{x=0}^{1}+2\int_0^{1}\frac{ \ln(1+1/x)}{1+x}\,{dx} \\& = \ln^2(2)+2\int_0^{1}\frac{ \ln(1+x)}{1+x}\,{dx}-2\int_0^{1}\frac{ \ln(x)}{1+x}\,{dx} \\& = \ln^2(2)+\ln^2(1+x)\bigg|_{x=0}^{1}-2\int_0^{1}\frac{\ln(x)}{1+x}\,{dx} \\& = 2\ln^2(2)-2 \int_0^{1}\frac{\ln(x)}{1+x}\,{dx}\end{aligned}$​

It remains to calculate the last integral. Let $x \mapsto 1-x$ which maps

$\displaystyle 2\int_0^{1}\frac{\ln(x)}{1+x}\,{dx} \mapsto \int_0^{1}\frac{\ln(1-x)}{x}\,{dx} $​

Then using the Maclaurin expansion of $\ln(1-x)$ we have

$\begin{aligned}2\int_0^{1}\frac{\ln(x)}{1+x}\,{dx} & = \int_0^{1}\frac{\ln(1-x)}{x}\,{dx} \\& = -\int_0^{1} \sum_{k \ge 0}\frac{x^{k}}{k+1} \,{dx} \\& = -\sum_{k \ge 0}\int_0^{1} \frac{x^{k}}{k+1}\,{dx} \\& = -\sum_{k \ge 0}\frac{1}{(k+1)^2} \\& = - \frac{\pi^2}{6}\end{aligned}$​

Therefore we have $\displaystyle \mathcal{I} = 2\ln^2(2)+\frac{\pi^2}{6}. $

Thankyou for your participation and a correct result, June29! Good job!(Happy)

Spoiler

Would you please explain the following step? Thankyou in advance!:

\[x \mapsto 1-x \Rightarrow 2\int_{0}^{1}\frac{\ln x}{1+x}dx \mapsto \int_{0}^{1}\frac{\ln (1-x)}{x}dx\]

I have a problem understanding the change of the denominator of the integrand:
- from $1+x$ to $x$. I´d expect: from $1+x$ to $2-x$?

 

[MountEvariste]

Thankyou for your participation and a correct result, June29! Good job!(Happy)

Spoiler

Would you please explain the following step? Thankyou in advance!:

\[x \mapsto 1-x \Rightarrow 2\int_{0}^{1}\frac{\ln x}{1+x}dx \mapsto \int_{0}^{1}\frac{\ln (1-x)}{x}dx\]

I have a problem understanding the change of the denominator of the integrand:
- from $1+x$ to $x$. I´d expect: from $1+x$ to $2-x$?

Spoiler

You're right! I wrote down the wrong map (Rofl) I've since spotted a different way by using

$\displaystyle \int_0^1 x^k \ln(x) \,{dx} = -\frac{1}{(1+k)^2}$ (which can be proven by integration by parts for example).

$\displaystyle \int_0^{1} \frac{\ln(x)}{1+x}\,{dx} = \int_0^{1}\ln(x)\sum_{k \ge 0}(-1)^kx^{k}\,{dx} = \sum_{k\ge0} (-1)^k\int_0^{1}x^k\ln(x)\,{dx} = -\sum_{k \ge 0}\frac{(-1)^k}{(1+k)^2}$

$= \displaystyle - \left(1-\frac{1}{2}\right)\sum_{k \ge 0} \frac{1}{(k+1)^2} = -\frac{\pi^2}{12}$ giving us $\displaystyle \mathcal{I} = 2\ln^2(2) -
2\left(-\frac{\pi^2}{12}\right) = 2\ln^2(2)+\frac{\pi^2}{6}.$

 

[lfdahl]

Spoiler

You're right! I wrote down the wrong map (Rofl) I've since spotted a different way by using

$\displaystyle \int_0^1 x^k \ln(x) \,{dx} = -\frac{1}{(1+k)^2}$ (which can be proven by integration by parts for example).

$\displaystyle \int_0^{1} \frac{\ln(x)}{1+x}\,{dx} = \int_0^{1}\ln(x)\sum_{k \ge 0}(-1)^kx^{k}\,{dx} = \sum_{k\ge0} (-1)^k\int_0^{1}x^k\ln(x)\,{dx} = -\sum_{k \ge 0}\frac{(-1)^k}{(1+k)^2}$

$= \displaystyle - \left(1-\frac{1}{2}\right)\sum_{k \ge 0} \frac{1}{(k+1)^2} = -\frac{\pi^2}{12}$ giving us $\displaystyle \mathcal{I} = 2\ln^2(2) -
2\left(-\frac{\pi^2}{12}\right) = 2\ln^2(2)+\frac{\pi^2}{6}.$

A fine solution path indeed! Thankyou June29!