# Integral $\int_0^1 \frac{x\log x+1-x}{x \log^2 x}\log(1+x)\, dx=\log\frac{4}{\pi}$

Hi I am trying to prove this $$I:=\int_{0}^{1} {x\log\left(\,x\,\right) + 1 - x \over x\log^{2}\left(\,x\,\right)}\, \log\left(\,1 + x\,\right)\,{\rm d}x=\log\left(\,4 \over \pi\,\right).$$ Thanks.

This is just a beautiful integral for many reasons. Logs are everywhere and an inspirational solution!!! I am not sure if breaking it up into three separate pieces is of any use, I tried that by writing $$I=\int_0^1\frac{ \log(1+x)}{\log x}dx+\int_0^1\frac{\log(1+x)}{x \log^2 x}dx-\int_0^1\frac{\log(1+x)}{\log^2 x}dx$$ but wasn't sure how to handle these. Also note that $$\int_0^1 \frac{x\log x+1-x}{x \log^2 x}dx=1,$$ in case that happened to come up anywhere along the calculation.

• Can you share where you are getting these? I know you say you used to compete in math competitions - can we see any of those resources? – Bennett Gardiner May 8 '14 at 4:34
• @BennettGardiner This is from Ryzhik's book on integrals (russian version). Others I post are from Demidovich - Problems in mathematical analysis (russian copy). A lot are also from Borwein papers. Many integrals I post are not from books and are just from my notes, so those resources were the city teachers & from competitions. The rest are from being an integral fanatic for 40 years. – Jeff Faraci May 8 '14 at 4:54
• @BennettGardiner Does this help? Thanks. – Jeff Faraci May 8 '14 at 4:55
• Jeff, please send me all of your integral resources. This is my email: ariawan.11@gmail.com. Thank you. :) – Tunk-Fey May 8 '14 at 5:03
• The three separate pieces you wrote down all diverge, so I'm pretty sure that this is not the way to go. Beautiful integral, though. – user111187 May 8 '14 at 6:46

Consider \begin{align*} \int_0^1 \frac{x\log{x}+1-x}{x}\, x^a\, \log{(1+x)}\, dx &= \int_0^1 \frac{x\log{x}+1-x}{x}\, x^a\, \sum_{n\ge 1} (-1)^{n+1}\frac{x^n}{n}\, dx\\ &=\sum_{n\ge 1} \int_0^1 \, (-1)^{n+1} (x\log{x}+1-x)\, \frac{x^{a+n-1}}{n}\, dx\\ &= \sum_{n\ge 1} - \frac{\left(-1\right)^{n+1}}{{\left(a + n + 1\right)}^{2} n} + \frac{\left(-1\right)^{n+1}}{{\left(a + n\right)} n} -\frac{\left(-1\right)^{n+1}}{{\left(a + n + 1\right)} n}\\ \int_0^1 \frac{x\log{x}+1-x}{x \log{x}}\, x^a\, \log{(1+x)}\, dx &= \sum_{n\ge 1} \frac{\left(-1\right)^{n + 1}}{n} \left(\frac{1}{a + n + 1} + \log\left(\frac{a + n}{a+n+1}\right)\right)\tag{$\int da$}\\ \int_0^1 \frac{x\log{x}+1-x}{x (\log{x})^2}\, x^a\, \log{(1+x)}\, dx &= \sum_{n\ge 1} \frac{\left(-1\right)^{n+1} {\left(a + n\right)} \log\left(\frac{a + n}{a + n + 1}\right) + \left(-1\right)^{n+1}}{n}\tag{$\int da$}\\ \end{align*} Subst. $a=0$ \begin{align*} \therefore \int_0^1 \frac{x\log{x}+1-x}{x (\log{x})^2}\, \log{(1+x)}\, dx &= \sum_{n\ge 1} \left(-1\right)^{n+1} \log\left(\frac{ n}{ n + 1}\right) + \frac{\left(-1\right)^{n+1}}{n}\hspace{20pt} \text{(Wallis product and log 2)}\\ &= \log{\left(\frac{2}{\pi}\right)}+\log{2} \\ &= \log{\left(\frac{4}{\pi}\right)} \end{align*}

• Nice, you found a more direct method. – user111187 May 9 '14 at 16:59
• Thanks.. I was hoping to get a general formula here also, but the summation looks difficult. – gar May 9 '14 at 17:33

Substituting $\log x = -u$ gives $$I = \int_0^{\infty} \frac{1-(x+1)e^{-x}}{x^2} \ln(1+e^{-x})$$ Noticing that $\frac{d}{dx} (e^{-x}-1)/x = \frac{1-(x+1)e^{-x}}{x^2}$, we integrate by parts to obtain $$I = -\ln 2- \int_0^{\infty} \frac{dx}{x}e^{-x} \frac{1-e^{-x}}{1+e^{-x}}$$ So the problem is reduced to showing that $$J:=\int_0^{\infty} \frac{dx}{x}e^{-x} \frac{1-e^{-x}}{1+e^{-x}}=\ln{\frac{\pi}{2}}$$ We have $$J = \sum_{k\geq0} (-1)^{k} \int_0^{\infty} \frac{dx}{x} \left( e^{-x}-e^{-2x}\right)e^{-kx}$$ (Here I used the geometric series and interchanged summation and integration.) $$= \sum_{k\geq0} (-1)^{k} \ln \frac{k+2}{k+1}$$ $$= \sum_{k\geq1} (-1)^{k-1} \ln \left( 1+\frac{1}{k} \right)$$ (Here I used a well-known integral that follows from Frullani's theorem)

$$= \sum_{k\geq1} \left[ \ln\left( 1+\frac{1}{2k-1}\right)-\ln\left( 1+\frac{1}{2k}\right) \right] \\ = \lim_{N\rightarrow\infty} \ln\left[ \prod_{k=0}^N \frac{1+\frac{1}{2k-1}}{1+\frac{1}{2k}} \right] \\ = \ln \left[ \lim_{N\rightarrow\infty} \prod_{k=0}^N \frac{4k^2}{4k^2-1} \right] \\ = -\ln \left[\prod_{k\geq0} \left( 1-\frac{1}{4k^2} \right) \right]\\ = -\ln \left[ \frac{\sin(\pi /2)} {\pi/2} \right]$$ Here I used the product formula for $\frac{\sin \pi x}{\pi x}$ Hence $$J= \ln \frac{\pi}{2}$$ as was to be proved.

• Very nice answer. +1 – Random Variable May 8 '14 at 16:26
• @RandomVariable Thank you, a big compliment coming from you! – user111187 May 8 '14 at 16:33
• This is very nice indeed! +1 – Pranav Arora May 8 '14 at 16:47
• @user111187: Can you please share a link about the product formula you used in the end? I managed to reach the final product (though my method isn't as elegant as yours) but didn't know how to evaluate the product. Thanks! – Pranav Arora May 8 '14 at 16:49
• @PranavArora A derivation of the formula that does not require complex analysis or advanced methods is given here: ocw.mit.edu/courses/mathematics/… – user111187 May 8 '14 at 16:53

Note $$\frac{1}{x\log^2x}=-\frac{d}{dx}\frac{1}{\log x}$$ and hence \begin{eqnarray} I&=&\int_{0}^{1} {x\log\left(\,x\,\right) + 1 - x \over x\log^{2}\left(\,x\,\right)}\, \log\left(\,1 + x\,\right)\,{\rm d}x\\ &=&-\int_{0}^{1} (x\log x + 1 - x)\log(1+x)d\frac{1}{\log x}\\ &=&-(x\log x + 1 - x)\log(1+x)\frac{1}{\log x}\bigg|_0^1+\int_{0}^{1}\frac{1}{\log x}d[ (x\log x + 1 - x)\log(1+x)]\\ &=&\int_0^1\frac{1}{\log x}\left(\frac{x \log x+1-x}{x+1}+\log x \log (x+1)\right)dx\\ &=&\int_0^1\log(x+1)dx+\int_0^1\frac{x}{x+1}+\int_0^1\frac{1}{\log x}\frac{1-x}{x+1}dx\\ &=&\log 2+J, \end{eqnarray} where $$J=\int_0^1\frac{1}{\log x}\frac{1-x}{x+1}dx.$$ Define $$f(a)=\int_0^1x^a\frac{1-x}{x+1}dx.$$ Then \begin{eqnarray} f(a)&=&\int_0^1\sum_{n=0}^\infty(-1)^nx^{a+n}(1-x)dx &=&\sum_{n=0}^\infty(-1)^n\left(\frac{1}{n+a+1}-\frac{1}{n+a+2}\right) \end{eqnarray} and hence $$\int_0^a f(a)da=\sum_{n=0}^\infty(-1)^n\log\frac{n+a+1}{n+a+2},$$ and $$J=\lim_{a\to0}\sum_{n=0}^\infty(-1)^n\log\frac{n+a+1}{n+a+2}=\sum_{n=0}^\infty(-1)^n\log\frac{n+1}{n+2}=\log\frac{2}{\pi}.$$ Thus $$I=\log\frac{4}{\pi}.$$