June29 said: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 said:Thankyou for your participation and a correct result, June29! Good job!(Happy)
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 $2-x$?
A fine solution path indeed! Thankyou June29!June29 said: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 $\displaystyle \mathcal{I} = 2\ln^2(2) -
2\left(-\frac{\pi^2}{12}\right) = 2\ln^2(2)+\frac{\pi^2}{6}.$
The integral challenge ∫ln2(1+x^(−1))dx is a mathematical problem that involves finding the indefinite integral of the function ln2(1+x^(−1)), which represents the natural logarithm of 2 multiplied by 1 plus the inverse of x.
To solve the integral challenge ∫ln2(1+x^(−1))dx, you can use the substitution method or integration by parts. The substitution method involves replacing the variable x with another variable, while integration by parts involves splitting the function into two parts and integrating each part separately.
The integral challenge ∫ln2(1+x^(−1))dx has significance in mathematics as it represents a type of integral that is commonly used in various applications, such as in physics and engineering. It also demonstrates the importance of understanding integration techniques in solving mathematical problems.
There are multiple methods that can be used to solve the integral challenge ∫ln2(1+x^(−1))dx, such as the substitution method, integration by parts, or using a table of integrals. The specific method used may vary depending on the complexity of the problem and the preference of the solver.
Yes, the integral challenge ∫ln2(1+x^(−1))dx can be solved by hand using various integration techniques. However, for more complex functions, it may be more efficient to use computer software or a calculator to find the solution.