Integrate $\int_0^1{\frac{\ln{x}}{1+x}}dx$ using $\sum{\frac{1}{k^2}}=\frac{\pi ^2}{6}$

Question

Integrate $$\int_0^1{\frac{\ln{x}}{1+x}}dx$$ using $$\sum{\frac{1}{k^2}}=\frac{\pi ^2}{6}$$

My Attempt:

I know that $$\lim_{n\to{\infty}}\sum_{r=0}^nf\bigg(\frac{r}{n}\bigg)\cdot\frac{1}{n}=\int_0^1f(x)dx$$

But I am not able to relate the given summation to the integral. How do I proceed? Any hints would be helpful.

Thank you.

Answer 1

Beside the possibility of using the subsitution $x=e^{-u}$ together with the properties of geometric series the integral can be evaluated by finding the anti-derivative directly in terms of the dilogarithm which is defined as

$$\operatorname{Li}_2(x)=\sum_{k=1}^{\infty}\frac{x^k}{k^2}\text{ and therefore respectively }\frac d{dx}\operatorname{Li}_2(x)=-\frac{\ln(1-x)}{x}$$

so the anti-derivative of the given integral can be expressed as

$$\int \frac{\ln x}{1+x}dx=\operatorname{Li}_2(-x)+\ln(x)\ln(1+x)+c$$

Plugging in the borders of integration leads to

$$\int_0^1 \frac{\ln x}{1+x}dx = [\operatorname{Li}_2(-x)+\ln(x)\ln(1+x)]_0^1=\operatorname{Li}_2(-1)$$

the last term - $\operatorname{Li}_2(-1)$ - can be evaluated by using your given series. Since the series we examine converge absolutely the following steps are allowed

$$\begin{align} \operatorname{Li}_2(-1)=\sum_{k=1}^{\infty}\frac{(-1)^k}{k^2}=-\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^2}&=-\left[\sum_{k=1}^{\infty}\frac1{k^2}-2\sum_{k=1}^{\infty}\frac1{(2k)^2}\right]\\ &=-\left[\sum_{k=1}^{\infty}\frac1{k^2}-\frac12\sum_{k=1}^{\infty}\frac1{k^2}\right]\\ &=-\left[\frac{\pi^2}6-\frac12\frac{\pi^2}6\right]\\ &=-\frac{\pi^2}{12} \end{align}$$

Answer 2

Expand the integrand in series: If $|x| < 1$, $(1+x)^{-1} = \sum_{n=0}^\infty (-x)^n$. Then you just need to figure out what $\int_0^1 x^n \ln x\,dx$ is (consider $ x = e^{-u}$) and sum the series.

Answer 3

Since the denominator is $1+x$, it reminds us that we can use the infinite geometric sequence to rewrite the integral as an infinite expansion$$\frac 1{1+x}=\sum\limits_{n\geq0}(-1)^n x^n$$Therefore$$\mathfrak{I}=\sum\limits_{n\geq0}(-1)^n\int\limits_0^1\mathrm dx\, x^n\log x=\sum\limits_{n\geq0}\frac {(-1)^{n+1}}{(n+1)^2}$$Observe that the last integral is evaluated using integration by parts. Make a change in index to see that$$\mathfrak{I}=\sum\limits_{n\geq1}\frac {(-1)^n}{n^2}=-\sum\limits_{n\geq1}\frac {(-1)^{n-1}}{n^2}$$This infinite series actually has a name: Dirichlet eta function and has a nice relation with $\zeta(2)$. Namely$$\eta(s)=(1-2^{1-s})\zeta(s)$$ Therefore, since $\zeta(2)=\frac {\pi^2}6$, it’s easy to see that$$\int\limits_0^1\mathrm dx\,\frac {\log x}{1+x}\color{blue}{=-\frac {\pi^2}{12}}$$

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EDIT: I forgot to add.$$\zeta(2)=\sum\limits_{k\geq1}\frac 1{k^2}=\frac {\pi^2}6$$

Answer 4

Here's one way of doing it (I think in the spirit of what you're asking for,using the fact that $- \sum_{i=1}^{\infty} \frac{ -1^i}{i^2} = \frac{\pi^2}{12}$) Consider the series expansion for $1/(1+x)$:

$$ \frac{1}{1+x} = \sum_{i=0}^{\infty} (-1)^i x^i $$

Let's first consider the indefinite integral

\begin{align*} \int \frac{ \ln x}{1+x} \,dx &=\int\sum_{i=0}^{\infty} (-1)^i x^i\ln x \,dx \\ &=\sum_{i=0}^{\infty} (-1)^i \int x^i \ln x \,dx \\&=\sum_{i=0}^{\infty} (-1)^i \left(\frac{x^{i+1}}{i+1} \ln x - \int \frac{x^i}{i+1} \,dx \right) + C \quad \text{integration by parts}\\ &= \sum_{i=0}^{\infty} (-1)^i \left(\frac{x^{i+1}}{i+1} \ln x - \frac{x^{i+1}}{(i+1)^2} \right) + C\\ \end{align*} Taking the definite integral over $[\epsilon, 1 - \epsilon]$ and letting $\epsilon$ go to $0$ gives our integral: $$ \int_0^1 \frac{\ln x}{1+x} \,dx = -\sum_{i=0}^{\infty} (-1)^i \frac{1}{(i+1)^2} = -\frac{\pi^2}{12} $$