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Evaluate $$\int^{1}_{1/2}\frac{\ln(x)}{1-x}dx$$

Try: Let $$ I =\int^{1}_{1/2}\frac{\ln x}{1-x}dx=\int^{1}_{1/2}\sum^{\infty}_{k=0}x^k\ln(x)dx$$

$$I =\sum^{\infty}_{k=0}\int^{1}_{1/2}x^k\ln(x)dx$$

Using integration by parts, gives: $$I =\sum^{\infty}_{k=0}\frac{\ln(x)x^{k+1}}{k+1}\bigg|^{1}_{1/2}-\sum^{\infty}_{k=0}\frac{x^{k+1}}{(k+1)^2}\bigg|^{1}_{1/2}$$

Can someone help me to solve it?

Zacky
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DXT
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4 Answers4

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Applying a substitutiong of $\,\large\frac{1-x}x=t \Rightarrow dx=-\frac{dt}{(1+t)^2}$ gives: $$\int_\frac12^1\frac{\ln x}{x\cdot\frac{1-x}{x}}dx=\int_0^1 \frac{\ln\left({\frac{1}{1+t}}\right)}{t(1+t)}dt$$ $$=\int_0^1 \frac{\ln(1+t)}{1+t}dt-\int_0^1 \frac{\ln(1+t)}{t}dt$$ $$=\frac{\ln^2(1+t)}{2}\bigg|_0^1 - \frac{\pi^2}{12}=\frac{\ln^2 2}{2}-\frac{\pi^2}{12}$$ The second integral can be found here.

Community
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Zacky
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Try the change of variable in your original integral: $1-x = y$. Then: $$I := \int_{1/2}^{1} \frac{\ln(x)}{1-x} \text{d}x = \int_{0}^{1/2} \frac{\ln(1-y)}{y} \text{d} y$$

Then you can use the series: $\ln(1-y) = - \sum_{n=1}^{\infty} \frac{y^n}{n}$.

Now you need to prove that you can permute the integral and the infinite sum. After that, it yields:

$$ I = - \sum_{n=0}^{\infty} \frac{1}{n+1} \int_{0}^{1/2} y^{n} \text{d}y$$

And I let you finish the computation. It is rather easy now.

Plussoyeur
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Here we will address the integral:

\begin{equation} I = \int_{\frac{1}{2}}^1 \frac{\ln(x)}{1 - x}\:dx \end{equation}

We first observe that:

\begin{equation} \frac{\partial}{\partial a} x^a = x^a \ln(x) \Longrightarrow \ln(x) = \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} x^a \end{equation}

Thus,

\begin{equation} I = \int_{\frac{1}{2}}^1 \frac{\ln(x)}{1 - x}\:dx = \int_{\frac{1}{2}}^1 \frac{\lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} x^a }{1 - x}\:dx \end{equation}

Which by Leibniz's Integral Rule becomes: \begin{equation} I = \int_{\frac{1}{2}}^1 \frac{\lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} x^a }{1 - x}\:dx = \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} \int_{\frac{1}{2}}^1 \frac{x^a}{1 - x}\:dx \end{equation}

We now employ the Taylor Series for $\frac{1}{1 - x}$ (which is convergent on the bounds of the integral):

\begin{equation} \frac{1}{1 - x} = \sum_{n = 0}^{\infty} x^n \end{equation}

Thus our integral becomes: \begin{align} I &= \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} \int_{\frac{1}{2}}^1 \frac{x^a}{1 - x}\:dx = \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} \int_{\frac{1}{2}}^1 x^a\sum_{n = 0}^{\infty} x^n \:dx = \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} \sum_{n = 0}^{\infty} \int_{\frac{1}{2}}^1 x^{a + n} \:dx \nonumber \\ &= \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} \sum_{n = 0}^{\infty} \left[ \frac{x^{a + n + 1}}{a + n + 1}\right]_{\frac{1}{2}}^1 = \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} \sum_{n = 0}^{\infty} \left[ \frac{1}{a + n + 1} - \frac{\left(\frac{1}{2}\right)^{a + n + 1}}{a + n + 1}\right] \nonumber \\ &= \lim_{a \rightarrow 0^+} \sum_{n = 0}^{\infty} \left[ \frac{-1}{\left(a + n + 1\right)^2} - \frac{\left(\frac{1}{2}\right)^{a + n + 1}}{\left(a + n + 1\right)^2} + \frac{\ln\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^{a + n + 1}}{a + n + 1}\right] \nonumber \\ &= \sum_{n = 0}^{\infty} \left[\frac{-1}{\left(n + 1\right)^2} - \frac{\left(\frac{1}{2}\right)^{n + 1}}{\left( n + 1\right)^2} + \frac{\ln\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^{n + 1}}{n + 1}\right] \nonumber \\ &= -\sum_{n = 0}^{\infty} \frac{1}{(n + 1)^2} - \frac{1}{2}\sum_{n = 0}^{\infty} \frac{\left(\frac{1}{2}\right)^n}{(n + 1)^2} + \frac{\ln\left(\frac{1}{2}\right)}{2}\sum_{n = 0}^{\infty} \frac{\left(\frac{1}{2}\right)^n}{n + 1} \nonumber \\ &= -\Phi\left(1,1,2 \right) - \frac{1}{2}\Phi\left(\frac{1}{2},1,2 \right) + \ln(\sqrt{2})\Phi\left(\frac{1}{2},1,1 \right) \end{align} Where $\Phi(\cdot, \cdot, \cdot, \cdot)$ is the Lerch Transcedent Function.

  • My solution is only of worth in situations where a substitution as per the answerer's is unknown or unable to be derived. –  Feb 25 '19 at 06:36
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\begin{align} \text{J}&=\int_{\frac{1}{2}}^{1}\frac{\ln x }{1-x}dx\\ &=\int_{0}^{1}\frac{\ln x }{1-x}dx-\int_0^{\frac{1}{2}}\frac{\ln x }{1-x}dx\\ &=\int_{0}^{1}\frac{\ln x }{1-x}dx-\left(\Big[-\ln(1-x)\ln x\Big]_0^{\frac{1}{2}}+\int_0^{\frac{1}{2}}\frac{\ln(1-x)}{x} \,dx\right)\\ &=\int_{0}^{1}\frac{\ln x }{1-x}dx+\ln^2 2-\int_0^{\frac{1}{2}}\frac{\ln(1-x)}{x} \,dx\\ \end{align}

In the latter integral perform the change of variable $y=1-x$,

\begin{align} J&=\int_{0}^{1}\frac{\ln x }{1-x}dx+\ln^2 2-\text{J}\\ \end{align}

Therefore,

\begin{align} J&=\frac{1}{2}\int_{0}^{1}\frac{\ln x }{1-x}dx+\frac{1}{2}\ln^2 2\\ &=\boxed{\frac{1}{2}\ln^2 2-\frac{\pi^2}{12}}\\ \end{align}

NB: I assume,

\begin{align} \int_{0}^{1}\frac{\ln x }{1-x}dx=-\frac{\pi^2}{6} \end{align}

FDP
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