# A math contest problem $\int_0^1\ln\left(1+\frac{\ln^2x}{4\,\pi^2}\right)\frac{\ln(1-x)}x \ \mathrm dx$

A friend of mine sent me a math contest problem that I am not able to solve (he does not know a solution either). So, I thought I might ask you for help.

Prove: $$\int_0^1\ln\left(1+\frac{\ln^2x}{4\,\pi^2}\right)\frac{\ln(1-x)}x dx=-\pi^2\left(4\,\zeta'(-1)+\frac23\right).$$

• @VladimirReshetnikopv, just curious - from what math contest was this problem taken? – Arnie Bebita-Dris Oct 12 '13 at 0:53
• @JoseArnaldoDris Somewhere in Russia, I do not know exactly where. – Vladimir Reshetnikov Oct 12 '13 at 15:32

Here is a solution: Let $I$ denote the integral. Then

\begin{align*} I &= - \int_{0}^{1} \log \left( 1 + \frac{\log^{2}x}{4\pi^{2}}\right) \mathrm{Li}_{2}'(x) \, dx \\ &= \left[ -\log \left( 1 + \frac{\log^{2}x}{4\pi^{2}}\right) \mathrm{Li}_{2}(x) \right]_{0}^{1} + 2 \int_{0}^{1} \frac{\log x}{4\pi^{2} + \log^{2} x} \frac{\mathrm{Li}_{2}(x)}{x} \, dx \\ &= -2 \int_{0}^{\infty} \frac{t}{4\pi^{2} + t^{2}} \mathrm{Li}_{2}(e^{-t}) \, dt \qquad (x = e^{-t}) \\ &= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \frac{t}{4\pi^{2} + t^{2}} \, e^{-nt} \, dt \\ &= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \left( \int_{0}^{\infty} \cos (2\pi u) e^{-tu} \, du \right) e^{-nt} \, dt \\ &= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \frac{\cos (2\pi u)}{u + n} \, du \\ &= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \frac{\cos (2\pi n u)}{u + 1} \, du \qquad (u \mapsto nu) \\ &= -2 \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{1} \frac{\cos (2\pi n u)}{u + k} \, du \\ &= -2 \sum_{k=1}^{\infty} \int_{0}^{1} \frac{1}{u + k} \left( \sum_{n=1}^{\infty} \frac{\cos (2\pi n u)}{n^{2}} \right) \, du. \end{align*}

Now we invoke the Fourier series of the Bernoulli polynomial $B_{2}(x)$:

$$\sum_{n=1}^{\infty} \frac{\cos 2\pi nx}{n^{2}} = \pi^{2} B_{2}(x) = \pi^{2} \left( x^{2} - x + \frac{1}{6} \right), \quad 0 \leq x \leq 1.$$

Then it follows that

\begin{align*} I &= -2\pi^{2} \sum_{k=1}^{\infty} \int_{0}^{1} \frac{u^{2} - u + \frac{1}{6}}{u + k} \, du \\ &= \pi^{2} \sum_{k=1}^{\infty} \left\{ 2k + 1 + 2\left( k^{2} + k + \frac{1}{6} \right) \log \left( \frac{k}{k+1} \right) \right\}. \end{align*}

Now we consider the exponential of the partial sum:

\begin{align*} &\exp \left[ \sum_{k=1}^{N} \left\{ 2k + 1 + 2\left( k^{2} + k + \frac{1}{6} \right) \log \left( \frac{k}{k+1} \right) \right\} \right] \\ &= e^{N^{2} + 2N} \prod_{k=1}^{N} \left( \frac{k}{k+1} \right)^{2k^{2} + 2k + \frac{1}{3}} \\ &= \frac{e^{N^{2} + 2N}}{(N + 1)^{2N^{2} + 2N + \frac{1}{3}}} \prod_{k=1}^{N} k^{4k} \\ &= \frac{e^{2N}}{\left(1 + \frac{1}{N}\right)^{2N^{2} + 2N + \frac{1}{3}}} \left\{ \frac{e^{N^{2}/4}}{N^{N^{2}/2 + N/2 + 1/12}} \prod_{k=1}^{N} k^{k} \right\}^{4}. \end{align*}

In view of the definition of Glaisher-Kinkelin constant $A$, we have

$$\lim_{N\to\infty} \exp \left[ \sum_{k=1}^{N} \left\{ 2k + 1 + 2\left( k^{2} + k + \frac{1}{6} \right) \log \left( \frac{k}{k+1} \right) \right\} \right] = \frac{A^{4}}{e}.$$

This, together with the identity $\log A = \frac{1}{12} - \zeta'(-1)$, yields

$$I = \pi^{2} ( 4 \log A - 1 ) = -\pi^{2} \left(4\zeta'(-1) + \frac{2}{3} \right)$$

as desired.

• How did you see this??? I am amazed. – Potato Oct 12 '13 at 2:25
• I am awed by the range of mathematical knowledge displayed here. Bravo! – Ron Gordon Oct 12 '13 at 13:26
• @sos440 It is amazing! I can understand that steps with dilogarithm in the beginning and Glaisher-Kinkelin constant at the end are obvious, but I cannot imagine how you filled everything in between in less than an hour. I think I will never be able to do integrals like this. – Vladimir Reshetnikov Oct 12 '13 at 15:39
• Thank you! At first I had no idea, so I attacked this with my favorite techniques. Then I found that the intermediate series resembled this one. So I applied the same technique and hopefully this approach yielded the answer. – Sangchul Lee Oct 12 '13 at 17:22
• @BennettGardiner, The underlying idea is the same as in the summation by parts. The product is of the form $$P = \prod_{k=1}^{N} \frac{k^{f(k)}}{(k+1)^{f(k)}}.$$ Now splitting the product into both denominator and numerator, $$P = \frac{\prod_{k=1}^{N} k^{f(k)}}{\prod_{k=1}^{N} (k+1)^{f(k)}} = \frac{\prod_{k=1}^{N} k^{f(k)}}{\prod_{k=2}^{N+1} k^{f(k-1)}}.$$ Merging the product again gives $$P = \frac{1}{(N+1)^{f(N)}} \prod_{k=1}^{N} k^{f(k) - f(k-1)}.$$ – Sangchul Lee Oct 29 '13 at 2:21

In this answer, I will provide another approach to proving that $$I=\int_0^1 \log\left(1+\frac{\log^2x}{4\pi^2}\right)\frac{\log(1-x)}{x}dx=-4\pi^2\zeta'(-1)-\frac{2\pi^2}{3}.$$

First, by using the functional equation for the zeta function, we will prove the following lemma:

Lemma 1: Let $$L=\int_{0}^{1}\log\left(\frac{\log^{2}(x)}{4\pi^{2}}\right)\frac{\log(1-x)}{x}dx.$$ Then $L=-4\pi^{2}\zeta'(-1)+\frac{\pi^{2}}{3}.$

Proof: Expanding the series for $\log(1-x)/x$, we obtain $$\int_{0}^{1}\log\left(\frac{\log^{2}(x)}{4\pi^{2}}\right)\frac{\log(1-x)}{x}dx=-\sum_{n=0}^{\infty}\int_{0}^{1}\log\left(\frac{\log^{2}(x)}{4\pi^{2}}\right)\frac{x^{n}}{n}\frac{dx}{x}.$$ Letting $x=e^{-2\pi t}$, this is $$-2\pi\sum_{n=0}^{\infty}\frac{1}{n}\int_{0}^{\infty}\log\left(t^{2}\right)e^{-2\pi tn}dt=-4\pi\sum_{n=0}^{\infty}\frac{1}{n}\mathcal{L}\left(\log x\right)(2\pi n),$$ where $\mathcal{L}$ denotes the Laplace transform. Since $\mathcal{L}(\log x)(2\pi n)=\frac{-1}{2\pi n}(\log(2\pi n)+\gamma)$, we arrive at $$L=2\sum_{n=0}^{\infty}\frac{1}{n^{2}}\left(\log(2\pi n)+\frac{\gamma}{n^{2}}\right)=2\left(\frac{\pi^{2}}{6}\log2\pi+\frac{\pi^{2}\gamma}{6}-\zeta^{'}(2)\right).$$ Using the derivative of the functional equation for the zeta function, it follows that $$L=-4\pi^{2}\zeta'(-1)+\frac{\pi^{2}}{3}.$$

Lemma 2: We have that $I-L=-\pi^2$.

Proof: As before, by expanding the series for $\frac{\log(1-x)}{x}$ and substituting $x=e^{-2\pi t}$ we have that $$I=-2\pi\sum_{n=0}^{\infty}\frac{1}{n}\int_{0}^{\infty}\log\left(1+t^{2}\right)e^{-2\pi tn}dt.$$ Since $$\int_{0}^{1}\frac{2x}{x^{2}+t^{2}}=\log\left(1+t^{2}\right)-\log(t^{2}),$$ it follows that $$-4\pi\sum_{n=0}^{\infty}\frac{1}{n}\int_{0}^{1}\int_{0}^{\infty}\frac{x}{x^{2}+t^{2}}e^{-2\pi tn}dtdx=I-L.$$ Using the fact that $\frac{x}{x^{2}+t^{2}}$ is the Laplace transform of $\cos$, we have that $$I-L=-4\pi\sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{1}\int_{0}^{\infty}\int_{0}^{\infty}\cos(tv)e^{-xv}e^{-2\pi nt}dvdtdx$$

$$=-4\pi\sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{1}\int_{0}^{\infty}\left(\int_{0}^{\infty}\cos(tv)e^{-2\pi nt}dt\right)e^{-xv}dvdx$$

$$=-4\pi\sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{1}\int_{0}^{\infty}\frac{2\pi n}{(2\pi n)^{2}+v^{2}}e^{-xv}dvdx$$

$$=-8\pi^{2}\int_{0}^{\infty}\left(\sum_{n=1}^{\infty}\frac{1}{(2\pi n)^{2}+v^{2}}\right)\left(\int_{0}^{1}e^{-xv}dx\right)dv.$$ Evaluating the integral in $x$, and using the cotangent identity $$\sum_{n=1}^{\infty}\frac{1}{\left(2\pi n\right)^{2}+v^{2}}=\frac{v\coth(v/2)-2}{4v^{2}},$$ we obtain $$I-L=-2\pi^{2}\int_{0}^{\infty}\left(\frac{v\coth(v/2)-2}{v^{2}}\right)\left(\frac{1-e^{-v}}{v}\right)dv.$$ Expanding out $\coth(v/2)$ in terms of exponentials, the integrand equals $\left(e^{-v}v+v+e^{-v}-2\right)/v^{3}$, which has a simple anti derivative. In particular $$\int_{0}^{\infty}\left(\frac{v\coth(v/2)-2}{v^{2}}\right)\left(\frac{1-e^{-v}}{v}\right)dv=-\frac{v+e^{-v}-1}{v^{2}}\biggr|_{v=0}^{v=\infty}=\frac{1}{2},$$ and so we have shown that $I-L=-\pi^{2}.$

Combining Lemma 1 and Lemma 2, it follows that $$I=-4\pi^2\zeta'(-1)-\frac{2\pi^2}{3},$$ as desired.

I realize it's been a while, but I thought this one looked interesting, so I thought I would give it a go.

Make the obvious sub $\displaystyle t=\frac{-\ln(x)}{2\pi}, \;\ x=e^{-2\pi x}, \;\ dx=-2\pi e^{-2\pi x}dx$

$\displaystyle2\pi\int_{0}^{\infty}\ln(t^{2}+1)\ln(1-e^{-2\pi t})dt$.

Use parts $\displaystyle u=\ln(1-e^{-2\pi t}), \;\ dv=\ln(t^{2}+1)dt, \;\ v=x\ln(t^{2}+1)+2(\tan^{-1}(t)-t), \;\ du=\frac{2\pi}{e^{2\pi t}-1}dt$

$\displaystyle -2\pi \left(2\pi\int_{0}^{\infty}\left(t\ln(t^{2}+1)+2\tan^{-1}(t)-2t\right)\cdot \frac{1}{e^{2\pi t}-1}\right)dt$

$=\displaystyle -4\pi^{2} \left(\int_{0}^{\infty}\frac{t\ln(t^{2}+1)}{e^{2\pi t}-1}dt+2\int_{0}^{\infty}\frac{\tan^{-1}(t)}{e^{2\pi t}-1}dt-2\int_{0}^{\infty}\frac{t}{e^{2\pi t}-1}dt\right)$

Now, the two leftmost integrals form a rather well-known identity that can be derived from Hermite's integral (among other ways).

$\displaystyle\int_{0}^{\infty}\frac{t\ln(t^{2}+1)}{e^{2\pi t}-1}dt+2\int_{0}^{\infty}\frac{\tan^{-1}(t)}{e^{2\pi t}-1}dt=\zeta'(-1)+1/4$

The integral on the right end is rather elementary and evaluates to $\displaystyle 2\int_{0}^{\infty}\frac{t}{e^{2\pi t}-1}dt=1/12$

Putting the pieces together results in $\displaystyle-4\pi^{2}\left(\zeta'(-1)+1/4-1/12\right)=-4\pi^{2}\zeta'(-1)-\frac{2\pi^{2}}{3}$

Hermite's integral

$\displaystyle\zeta (a,t)=\sum_{n=0}^{\infty}\frac{1}{(n+t)^{a}}=\frac{t^{-a}}{2}+\frac{t^{1-a}}{a-1}+i\int_{0}^{\infty}\frac{(t+ix)^{-a}-(t-ix)^{-a}}{e^{2\pi x}-1}dx$

Diffing this w.r.t a leads to

$\displaystyle \zeta'(a,t)=\frac{-t^{-a}}{2}log(t)-\frac{t^{1-a}(1+(a-1)log(t))}{(a-1)^{2}}+i\int_{0}^{\infty}\frac{(t-ix)^{-a}log(t-ix)-(t+ix)^{-a}log(t+ix)}{e^{2\pi x}-1}dx$

By letting $a=-1, \;\ t=1$ leads to the two log/arctan integrals in question.

If we let $x\mapsto e^{-x}$ and then use the series of $\log(1-e^{-x})$, we obtain that

$$\underbrace{\int_0^{\infty} \sum_{n=1}^{\infty}\log(4 \pi^2) \frac{e^{-y n}}{n} \ dy}_{\displaystyle \zeta(2) \log(4 \pi^2)} -\underbrace{\sum_{n=1}^{\infty}\int_0^{\infty} \log(4 \pi^2+y^2) \frac{e^{-y n}}{n} \ dy}_{\displaystyle -2\sum_{n=1}^{\infty}\frac{\operatorname{Ci}(2\pi n)}{n^2}+2\log(2)\zeta(2)+2\log(\pi)\zeta(2)}=\pi^{2} ( 4 \log A - 1 )$$

where in the last integral I used the exponential integral and $1.22a$ from this paper.

Q.E.D.