# An alternating Euler Sum

In this post, we will evaluate the famous Euler sum: $$\sum_{n=1}^\infty (-1)^{n+1}\frac{H_n}{n^3}\quad \color{blue}{\cdots (*)}$$ where $H_n = \sum_{k=1}^n \frac{1}{k}=\int_0^1 \frac{1-t^n}{1-t}dt$ is the $n$-th harmonic number. This series resists contour integration techniques which makes it’s computation quite challenging. We will start with the following well known generating function identity: $$\sum_{n=1}^\infty H_n t^n = -\frac{\log(1-t)}{1-t} ,\quad -1 \leq t < 1$$ Dividing the above equation by $t$ and integrating both sides, gives us: \begin{aligned} \sum_{n=1}^\infty H_n \frac{x^n}{n} &= -\int_0^x\frac{\log(1-t)}{t(1-t)}dt \\ &= -\int_0^x \left(\frac{1}{t} + \frac{1}{1-t} \right)\log(1-t) \; dt \\ &= \text{Li}_2(x) + \frac{1}{2}\log^2(1-x) \end{aligned} where $-1\leq x < 1$.

## 1. Computation of $\sum_{n=1}^\infty \frac{H_n}{n^2}x^n$

Now, we will use the same process to evaluate $\sum_{n=1}^\infty \frac{H_n}{n^2}x^n$ but we’ll have to divide the evaluation into two parts.

(I) First consider the case when $0 \leq x < 1$. Actually, we don’t need this case for evaluating (*) but I’ll do it anyway for completeness. We have: \begin{aligned} \sum_{n=1}^\infty \frac{H_n}{n^2} x^n &= \int_0^x\left(\frac{\text{Li}_2(t)}{t} + \frac{1}{2}\frac{\log^2(1-t)}{t}\right)dt \\ &= \text{Li}_3(x) + \frac{1}{2}\int_0^x \frac{\log^2(1-t)}{t} dt \\ &= \text{Li}_3(x) + \frac{1}{2}\int_{\log(1-x)}^0 \frac{t^2 e^t}{1-e^t}dt \quad (t\mapsto 1-e^t) \\ &= \text{Li}_3(x) + \frac{1}{2}\sum_{n=1}^\infty \int_{\log(1-x)}^0 t^2 e^{nt} \; dt \\ &= \text{Li}_3(x) + \frac{1}{2}\sum_{n=1}^\infty \left(\frac{2}{n^3} -\frac{\log^2(1-x)}{n}(1-x)^n+\frac{2\log(1-x)}{n^2}(1-x)^n-2\frac{(1-x)^n}{n^3}\right) \\ &= \text{Li}_3(x) + \frac{\log^2(1-x)\log(x)}{2} + \log(1-x)\text{Li}_2(1-x) - \text{Li}_3(1-x) + \zeta(3) \quad\color{blue}{\cdots (1)} \end{aligned} (II) Similarly, for the case $-1\leq x < 0$, we obtain:

\begin{aligned} \sum_{n=1}^\infty \frac{H_n}{n^2}x^n &= \text{Li}_3(x) -\frac{1}{2}\int_x^0 \frac{\log^2(1-t)}{t}dt \\ &= \text{Li}_3(x) +\frac{1}{2}\int_{0}^{-x} \frac{\log^2(1+t)}{t}dt \quad (t\mapsto -t) \\ &= \text{Li}_3(x) +\frac{1}{2}\int_{0}^{\log(1-x)} \frac{t^2 e^t}{e^t-1}dt \quad (t\mapsto e^t-1) \\ &= \text{Li}_3(x) +\frac{1}{2} \sum_{n=0}^\infty \int_0^{\log(1-x)} t^2 e^{-nt} dt \\ &= \text{Li}_3(x) +\frac{\log^3(1-x)}{6} + \frac{1}{2}\sum_{n=1}^\infty \int_0^{\log(1-x)} t^2 e^{-nt} dt \\ &= \text{Li}_3(x) +\frac{\log^3(1-x)}{6} + \frac{1}{2}\sum_{n=1}^\infty \left(\frac{2}{n^3} -\log^2(1-x)\frac{(1-x)^{-n}}{n}-2\log(1-x)\frac{(1-x)^{-n}}{n^2}-2\frac{(1-x)^{-n}}{n^3}\right) \\ &= \text{Li}_3(x) -\frac{\log^3(1-x)}{3} +\frac{\log^2(1-x)\log(-x)}{2} -\log(1-x)\text{Li}_2\left( \frac{1}{1-x}\right)-\text{Li}_3\left(\frac{1}{1-x}\right)+\zeta(3) \\ &\quad\color{blue}{\cdots (2)} \end{aligned}

## 2. Evaluation of $\sum_{n=1}^\infty \frac{H_n}{n^3}x^n$

This is the hardest step in the evaluation. Once again, we will divide the evaluation into two parts.

(I) Like section 1, we’ll first consider the case when $0\leq x < 1$.

Using integration by parts, we obtain: \begin{aligned} \int_0^x \frac{\zeta(3)-\text{Li}_3(1-t)}{t}dt &= \log(t)\left(\zeta(3)-\text{Li}_3(1-t) \right) \Big|_{0}^x - \int_0^x \frac{\log(t)\text{Li}_2(1-t)}{1-t}dt \\ &= \log(x)\left(\zeta(3)-\text{Li}_3(1-x) \right) - \frac{1}{2}\text{Li}_2^2 (1-x) + \frac{\zeta^2 (2)}{2} \quad \color{blue}{\cdots (3)} \end{aligned}

\begin{aligned} \int_0^x \frac{\log(1-t)\text{Li}_2(1-t)}{t}dt &= \log(t)\log(1-t)\text{Li}_2(1-t) \Big|_0^x - \int_0^x \log(t) \left(\frac{-\text{Li}_2(1-t)}{1-t} + \frac{\log(1-t)\log(t)}{1-t}\right)dt \\ &= \log(x)\log(1-x)\text{Li}_2(1-x) + \frac{1}{2}\text{Li}_2^2(1-x)-\frac{\zeta^2(2)}{2} - \int_0^x \frac{\log^2(t) \log(1-t)}{1-t}dt \\ &\quad \color{blue}{\cdots (4)} \end{aligned}

\begin{aligned} \int_0^x \frac{\log(t)\log^2(1-t)}{2t}dt &= \frac{1}{4}\log^2(t)\log^2(1-t)\Big|_0^x +\frac{1}{2}\int_0^x \frac{\log^2(t)\log(1-t)}{1-t}dt \\ &= \frac{1}{4}\log^2(x)\log^2(1-x) +\frac{1}{2}\int_0^x \frac{\log^2(t)\log(1-t)}{1-t}dt \quad \color{blue}{\cdots (5)} \end{aligned}

Putting together the results of equations (3), (4) and (5) gives us: \begin{aligned} \sum_{n=1}^\infty \frac{H_n}{n^3}x^n &= \text{Li}_4(x) +\frac{1}{4}\log^2(x)\log^2(1-x) + \log(x)\log(1-x)\text{Li}_2(1-x)\\ &\quad + \log(x)\left(\zeta(3)-\text{Li}_3(1-x) \right) -\frac{1}{2}\int_0^x \frac{\log^2(t)\log(1-t)}{1-t}dt \quad \color{blue}{\cdots (6)} \end{aligned}

(II) Now, let $-1\leq x < 0$. We have: \begin{aligned} -\int_x^0 \frac{\zeta(3)-\text{Li}_3\left(\frac{1}{1-t}\right)}{t}dt &= \int_0^{\frac{-x}{1-x}}\frac{\zeta(3)-\text{Li}_3(1-t)}{t(1-t)}dt \quad \left(t\mapsto\frac{-t}{1-t} \right)\\ &= \int_0^{\frac{-x}{1-x}}\frac{\zeta(3)-\text{Li}_3(1-t)}{t}dt + \int_0^{\frac{-x}{1-x}}\frac{\zeta(3)-\text{Li}_3(1-t)}{1-t}dt \\ &= \log\left(\frac{-x}{1-x}\right)\left(\zeta(3) - \text{Li}_3\left(\frac{1}{1-x}\right) \right) - \frac{1}{2}\text{Li}_2^2\left(\frac{1}{1-x}\right) + \frac{\zeta^2(2)}{2} \\ &\quad + \zeta(3)\log(1-x) + \text{Li}_4\left(\frac{1}{1-x} \right) - \zeta(4) \quad \color{blue}{\cdots (7)} \end{aligned} Here, we used equation (3) to simplify the integral after the transformation. Similarly, we obtain: \begin{aligned} \int_x^0 \frac{\log(1-t)\text{Li}_2\left(\frac{1}{1-t}\right)}{t}dt &= \int_0^{\frac{-x}{1-x}}\frac{\log(1-t)\text{Li}_2(1-t)}{t(1-t)}dt \\ &= \int_0^{\frac{-x}{1-x}}\frac{\log(1-t)\text{Li}_2(1-t)}{t}dt + \int_0^{\frac{-x}{1-x}}\frac{\log(1-t)\text{Li}_2(1-t)}{1-t}dt \\ &= -\log\left(\frac{-x}{1-x} \right)\log(1-x)\text{Li}_2\left(\frac{1}{1-x}\right)+\frac{1}{2}\text{Li}_2^2\left(\frac{1}{1-x}\right)-\frac{\zeta^2(2)}{2} \\ &\quad +\text{Li}_4\left(\frac{1}{1-x}\right)+\log(1-x)\text{Li}_3\left(\frac{1}{1-x}\right) -\zeta(4) \\ &\quad -\int_0^{\frac{-x}{1-x}}\frac{\log^2(t)\log(1-t)}{1-t}dt \quad \color{blue}{\cdots (8)} \end{aligned}

\begin{aligned} \int_x^0 \left( \frac{\log^3(1-t)}{3t} - \frac{\log^2(1-t)\log(-t)}{2t}\right) dt &= \frac{1}{6}\log^3(1-x)\log\left(\frac{-x}{1-x}\right) + \frac{1}{24}\log^4(1-x) \\ &\quad +\frac{1}{4}\log^2\left(\frac{-x}{1-x}\right)\log^2(1-x) + \frac{1}{2}\int_0^{\frac{-x}{1-x}}\frac{\log^2(t)\log(1-t)}{1-t}dt \\ &\quad \color{blue}{\cdots (9)} \end{aligned}

Finally, putting these results together gives:

\begin{aligned} \sum_{n=1}^\infty \frac{H_n}{n^3}x^n &= \text{Li}_4(x) + 2\text{Li}_4\left( \frac{1}{1-x}\right) + \log\left(\frac{-x}{1-x} \right)\left(-\text{Li}_3\left(\frac{1}{1-x}\right) +\frac{\log^3(1-x)}{6}-\log(1-x)\text{Li}_2\left(\frac{1}{1-x} \right)\right) \\ &\quad + \log(1-x)\text{Li}_3\left(\frac{1}{1-x}\right) + \frac{\log^4(1-x)}{24} +\frac{1}{4}\log^2 \left(\frac{-x}{1-x}\right)\log^2(1-x) +\zeta(3) \log(-x)- 2\zeta(4) \\ &\quad -\frac{1}{2}\int_0^{\frac{-x}{1-x}}\frac{\log^2(t)\log(1-t)}{1-t}dt \quad \color{blue}{\cdots (10)} \end{aligned}

The integral $\int \frac{\log^2(t)\log(1-t)}{1-t}dt$ can be evaluated in terms of Polylogarithms (refer section 7.6 of “Polylogarithms and Associated Functions” by Leonard Lewin). But it turns out that for evaluating (*) such a calculation is not needed.

## 3. Evaluation of $\int_0^{\frac{1}{2}}\frac{\log^2(t)\log(1-t)}{1-t}dt$

Let $I=\int_0^{1}\frac{\log^2(t)\log(1-t)}{1-t}dt$ and $J=\int_0^{\frac{1}{2}}\frac{\log^2(t)\log(1-t)}{1-t}dt$. The integral, $I$, can be evaluated by the differentiation under the integral technique. \begin{aligned} I &= \lim_{y\to 0^+}\lim_{x\to 1}\frac{\partial^2 }{\partial x^2} \frac{\partial }{\partial y} \int_0^1 t^{x-1} (1-t)^{y-1} dt \\ &= \lim_{y\to 0^+}\lim_{x\to 1}\frac{\partial^2 }{\partial x^2} \frac{\partial }{\partial y} \left\{\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} \right\} \end{aligned} It takes a bit of effort to compute the derivatives in terms of Polygamma functions so I won’t write the details here. One can verify that the end result is: $$I = -\frac{\pi^4}{180}$$ Now, apply integration by parts to the integral $J$: \begin{aligned} J &= \int_0^{\frac{1}{2}}\frac{\log^2(t)\log(1-t)}{1-t}dt \\ &= -\frac{\log^2(t)\log^2(1-t)}{2}\Big|_0^{\frac{1}{2}} + \int_0^{\frac{1}{2}} \frac{\log(t)\log^2(1-t)}{t}dt \\ &= -\frac{\log^4(2)}{2} + \int_{\frac{1}{2}}^1 \frac{\log^2(t)\log(1-t)}{1-t}dt \quad (t\mapsto 1-t)\\ &= -\frac{\log^4(4)}{2} + I-J \end{aligned} Now, one can solve the above equation for $J$ to get: \begin{aligned} J &= -\frac{\log^4(2)}{4}+\frac{I}{2} \\ &= -\frac{\log^4(2)}{4}-\frac{\pi^4}{360} \end{aligned}

Finally, we have every thing needed to evaluate (*). Plugging in $x=-1$ in equation (10) and performing some simplifications gives us the beautiful result: $$\boxed{\sum_{n=1}^\infty \frac{H_n}{n^3}(-1)^{n+1} = \frac{11\pi^4}{360}+\frac{\pi^2}{12}\log^2(2)-\frac{\log^4(2)}{12}-\frac{7}{4}\log(2)\zeta(3) -2\text{Li}_4\left(\frac{1}{2}\right)}$$ A similar result is obtained by plugging $x=\frac{1}{2}$ into equation (6): $$\boxed{\sum_{n=1}^\infty \frac{H_n}{n^3 2^n}= \frac{\pi^4}{720}+\frac{\log^4(2)}{24}-\frac{1}{8}\log(2)\zeta(3) +\text{Li}_4\left(\frac{1}{2}\right)}$$

## 4. Further Results

Let $H_n^{(2)} = \zeta(2) - \psi_1(n+1) = \sum_{k=1}^n \frac{1}{k^2}$. Then, it is easy to verify that: $$\sum_{n=1}^\infty t^n H_n^{(2)} = \frac{\text{Li}_2(t)}{1-t}, \quad -1\leq t < 1$$ Now, dividing the above equation by $t$ and integrating both sides, gives us: \begin{aligned} \sum_{n=1}^\infty \frac{H_n^{(2)}}{n}x^n &= \int_0^x \text{Li}_2(t)\left( \frac{1}{t}+\frac{1}{1-t}\right) \\ &= \text{Li}_3(x) + \int_0^x \frac{\text{Li}_2(t)}{1-t} dt \\ &= \text{Li}_3(x) -\log(1-x)\text{Li}_2(x) - \int_0^x \frac{\log^2(1-t)}{t}dt \end{aligned} From our previous calculation, we know that $$\int_0^x \frac{\log^2(1-t)}{t}dt = 2\sum_{n=1}^\infty \frac{H_n}{n^2}x^n -2\text{Li}_3(x)$$ Hence, we obtain the following relation: $$\sum_{n=1}^\infty \frac{H_n^{(2)}}{n}x^n =3\text{Li}_3(x)-\log(1-x)\text{Li}_2(x) - 2\sum_{n=1}^\infty \frac{H_n}{n^2}x^n \quad \color{blue}{\cdots (11)}$$ Once again, divide the above equation by $x$ and integrate both sides to get: $$\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^2}x^n = 3\text{Li}_4(x) + \frac{1}{2}\text{Li}_2^2(x) - 2\sum_{n=1}^\infty \frac{H_n}{n^3}x^n \quad \color{blue}{\cdots (12)}$$ Equations (11) and (12) are valid for $-1\leq x < 1$. Plugging in $x=\frac{1}{2}$ in equation (12) and using the known closed form for $\sum_{n=1}^\infty \frac{H_n}{n^3 2^n}$ gives us: $$\boxed{ \sum_{n=1}^\infty \frac{H_n^{(2)}}{n^2 2^n} = \frac{\pi^4}{1440}-\frac{\pi^2}{24}\log^2(2) + \frac{\log^4(2)}{24}+ \frac{1}{4}\log(2)\zeta(3) + \text{Li}_4\left(\frac{1}{2}\right) }$$

# Ramanujan’s Master Theorem: Part I

Ramanujan’s master theorem (RMT) is powerful tool that provides the Mellin transform of an analytic function using it’s power series around zero.

Theorem (RMT): If the complex-valued function $f(z)$ has the following expansion $$f(z) = \sum_{n=0}^\infty \frac{\varphi(n)}{n!}(-z)^n$$ then it’s Mellin transform is given by $$\int_0^\infty x^{s-1} f(x)\; dx = \Gamma(s) \varphi(-s)$$

Sometimes RMT provides short proofs for integrals that appear to be very difficult. One such example is the following: $$\int_0^\infty x^{\sigma-1} J_\nu (x) \; dx = \frac{2^{\sigma-1}\Gamma\left(\frac{\sigma+\nu}{2}\right)}{\Gamma\left(\frac{2-\sigma+\nu}{2} \right)} \; ,\quad -\nu < \sigma < \frac{3}{2} \quad (1)$$ where $J_\nu (x)$ is the Bessel function of the first kind.

Indeed, $J_\nu(x)$ has the following series expansion around $x=0$: $$J_\nu(x) =\sum_{n=0}^\infty \frac{(-1)^n }{n! \Gamma(n+\nu+1)}\left(\frac{x}{2} \right)^{\nu+2n}$$ Hence, we may apply the RMT to the function $x^{-\nu/2}J_\nu(2\sqrt{x})$ to obtain: $$\int_0^\infty x^{s-\nu/2-1} J_\nu(2\sqrt{x}) dx=\frac{\Gamma(s)}{\Gamma(\nu+1-s)}$$ Now, one can perform the substitutions $x=t^2/4$ and $s=\frac{\sigma+\nu}{2}$ to transform the above integral into the integral of equation (1).

It turns out that a similar approach also works for the product $J_\nu (x) J_\mu(x)$. First, we need to compute it’s series expansion. This can be done by multiplying the individual series expansions and rearranging/grouping certain terms.

\begin{aligned} J_{\nu}(x)J_{\mu}(x) &= \left(\sum_{n=0}^\infty \frac{(-1)^n }{n! \Gamma(n+\nu+1)}\left(\frac{x}{2} \right)^{\nu+2n}\right) \left(\sum_{m=0}^\infty \frac{(-1)^m }{m! \Gamma(m+\nu+1)}\left(\frac{x}{2} \right)^{\nu+2m} \right) \\ &= \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{(-1)^{n+m} x^{\nu+\mu+2(m+n)}}{2^{\nu+\mu+2(m+n)}} \cdot \frac{1}{m! n!\Gamma(n+\nu+1)\Gamma(m+\mu+1)} \\ &= \sum_{r=0}^\infty \frac{(-1)^{r} x^{\nu+\mu+2r}}{2^{\nu+\mu+2r}}\sum\limits_{\substack{0\leq m,n \leq r \\ m+n=r}}\frac{1}{m! n!\Gamma(n+\nu+1)\Gamma(m+\mu+1)} \\ &= \sum_{r=0}^\infty \frac{(-1)^{r} x^{\nu+\mu+2r}}{2^{\nu+\mu+2r} r! \Gamma(r+\nu+1)\Gamma(r+\mu+1)}\sum\limits_{\substack{0\leq m,n \leq r \\ m+n=r}}\binom{r}{m,n} \left(\prod_{i=m+1}^r (\nu+i)\right) \left(\prod_{j=n+1}^r (\mu+j)\right) \\ &= \sum_{r=0}^\infty \frac{(-1)^{r} x^{\nu+\mu+2r}}{2^{\nu+\mu+2r} r! \Gamma(r+\nu+1)\Gamma(r+\mu+1)}\sum _{n=0}^r\binom{r}{n}(\nu +r)^{\underline{n}}(\mu +r)^{\underline{r-n}} \\ &= \sum_{r=0}^\infty \frac{(-1)^{r} x^{\nu+\mu+2r}}{2^{\nu+\mu+2r} r! \Gamma(r+\nu+1)\Gamma(r+\mu+1)} (\nu +\mu+2r)^{\underline{r}} \\ &= \sum_{r=0}^\infty \frac{(-1)^{r} \Gamma(\nu+\mu+2r+1)}{ r! \Gamma(r+\nu+1)\Gamma(r+\mu+1)\Gamma(\mu+\nu+r+1)} \left( \frac{x}{2}\right)^{\nu+\mu+2r} \quad \; \quad (2) \end{aligned}

To simplify the inner summation, we used the binomial theorem for falling factorials.

Therefore, applying RMT to the function $x^{-\frac{\nu+\mu}{2}}J_{\nu}(2\sqrt{x})J_{\mu}(2\sqrt{x})$, gives us: $$\int_0^\infty x^{s-\frac{\nu+\mu}{2}-1} J_{\nu}(2\sqrt{x})J_{\mu}(2\sqrt{x}) \; dx = \frac{\Gamma(s)\Gamma(\nu+\mu+1-2s)}{\Gamma(\nu+1-s)\Gamma(\mu+1-s)\Gamma(\nu+\mu+1-s)}$$ Finally, make the substitutions $x=t^2/4$ and $s=\frac{\sigma+\nu+\mu}{2}$ in the above integral to get: $$\boxed{ \int_0^\infty x^{\sigma-1}J_\nu(x)J_\mu(x)\; dx = \frac{2^{\sigma-1} \Gamma\left(\frac{\nu+\mu+\sigma}{2} \right)\Gamma\left(1-\sigma\right)}{\Gamma\left(\frac{\nu-\mu-\sigma}{2} +1\right)\Gamma\left(\frac{\mu-\nu-\sigma}{2} +1\right)\Gamma\left(\frac{\nu+\mu-\sigma}{2} + 1\right)} } \quad \;\quad (3)$$ The above equation holds provided that the condition $-(\nu+\mu)< \sigma < 1$ is satisfied.

# Fourier Series of the Log-Gamma Function and Vardi’s Integral

In this post, we will compute the Fourier series expansion of the Log-Gamma function and use it to prove the beautiful Vardi’s integral: $$\int_0^1 \frac{\log\log \left(\frac{1}{x}\right)}{1+x^2}dx = \frac{\pi}{2}\log\left(\sqrt{2\pi}\frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)} \right)$$

Fourier Series of the Log-Gamma Function:

For $s\in (0,1)$, we have $$\log \Gamma(s) = \left(\frac{1}{2}-s \right)(\gamma + \log 2)-\frac{1}{2}\log(\sin (\pi s)) + (1-s) \log(\pi) + \frac{1}{\pi}\sum_{n=1}^\infty \frac{\sin(2\pi n s)\log n}{n} \quad \color{blue}{(1)}$$

Proof: It suffices to do the following integrals: \begin{aligned} \int_0^1 \log \Gamma (s) \; ds &= \frac{1}{2}\log(2\pi) \\ \int_0^1 \log \Gamma(s) \cos(2\pi n s)\; ds &= \frac{1}{4n} \quad \forall n\in \mathbb{Z}^+ \\ \int_0^1 \log \Gamma(s) \sin(2\pi n s)\; ds &= \frac{\gamma + \log(2n\pi )}{2n\pi}\quad \forall n\in \mathbb{Z}^+ \end{aligned}

The first integral can be evaluated with the help of Euler’s reflection formula: \begin{aligned} \int_0^1 \log \Gamma(s) \; ds &= \frac{1}{2} \int_0^1 \log \Gamma(s) \; ds + \frac{1}{2}\int_0^1 \log \Gamma(1-s) \; ds \\ &= \frac{1}{2}\int_0^1 \log\left(\frac{\pi}{\sin(\pi s)} \right)\; ds \\ &= \frac{\log (\pi)}{2} - \frac{1}{2}\int_0^1 \log(\sin(\pi s))\; ds \\ &= \frac{\log(2\pi)}{2} \end{aligned}

The second integral can also be dealt in a similar way. We have: \begin{aligned} \int_0^1 \log \Gamma(s) \cos(2\pi n s)\; ds &= \frac{1}{2}\int_0^1 \log \Gamma(s) \cos(2\pi n s)\; ds + \frac{1}{2}\int_0^1 \log \Gamma(1-s) \cos(2\pi n (1-s))\; ds \\ &= \frac{1}{2}\int_0^1 \log\left(\frac{\pi}{\sin(\pi s)} \right)\cos(2\pi n s)\; ds \\ &= -\frac{1}{2}\int_0^1 \log(2\sin(\pi s))\cos(2\pi n s)\; ds \\ &= \frac{1}{2}\int_0^1 \left(\sum_{k=1}^\infty \frac{\cos(2\pi k s)}{k} \right)\cos(2\pi n s)\; ds \\ &= \frac{1}{2}\sum_{k=1}^\infty \frac{1}{k}\int_0^1 \cos(2\pi k s)\cos(2\pi n s)\; ds \\ &= \frac{1}{4n} \end{aligned} In the above calculation, we used the well known Fourier series: $$\log(2\sin (\pi s)) = -\sum_{k=1}^\infty \frac{\cos(2\pi k s)}{k} \quad \forall s\in (0,1) \quad\quad \color{blue}{(2)}$$

The third integral is the most troublesome of all since the trick involving Euler’s reflection formula does not work. We will instead use the following integral representation of the Log-Gamma function: $$\log \Gamma(s) = \int_0^\infty \left(\frac{s-1}{t e^t} -\frac{1-e^{t(1-s)}}{t(e^t-1)}\right) dt \quad \forall s>0$$ One can easily verify the above equation via the differentiation under the integral technique. Therefore, upon changing the order of integration we get: \begin{aligned} \int_0^1 \log \Gamma(s) \sin(2\pi n s)\; ds &= \int_0^\infty \left(\frac{1}{t e^t}\int_0^1 s\sin(2\pi ns)\; ds +\frac{1}{t(e^t-1)}\int_0^1e^{t(1-s)}\sin(2\pi n s)\; ds \right)dt \\ &= \int_0^\infty \left[\frac{1}{t e^t}\left(-\frac{1}{2\pi n}\right) +\frac{1}{t(e^t-1)}\left(\frac{2\pi n (e^t-1)}{t^2 + (2\pi n)^2} \right) \right]dt \\ &= \int_0^\infty \left(-\frac{1}{2\pi n t e^t} + \frac{2\pi n}{t\left( t^2 + (2\pi n)^2 \right)} \right)dt \\ &= \frac{1}{2\pi n}\int_0^\infty\left(-\frac{1}{te^t} +\frac{1}{t}-\frac{t}{t^2 + (2\pi n)^2} \right) dt \\ &= \frac{1}{2\pi n}\left(\gamma + \left[ \log t - \log (t) e^{-t} - \frac{1}{2}\log\left( t^2 + (2\pi n)^2 \right)\right]_{t=0}^\infty \right) \\ &= \frac{\gamma + \log(2\pi n)}{2\pi n} \end{aligned}

To get equation (1), one needs to piece together all these calculations. Equation (2) along with the result: $$\pi \left(\frac{1}{2}- s\right) = \sum_{k=1}^\infty \frac{\sin(2\pi k s)}{k} \quad \forall s\in (0,1)$$ are needed to perform simplifications.

Now, we turn our attention to the Vardi’s integral: \begin{aligned} I &= \int_0^1 \frac{\log\log \left(\frac{1}{x}\right)}{1+x^2}dx \\ &= \int_0^\infty \frac{\log(t) e^{-t}}{1 + e^{-2t}}dt \\ &= \int_0^\infty \log(t)\sum_{n=0}^\infty (-1)^n e^{-(2n+1)t} \; dt\\ &= \sum_{n=0}^\infty (-1)^n \int_0^\infty \log (t) e^{-(2n+1)t}\; dt \\ &= -\sum_{n=0}^\infty (-1)^n \left(\frac{\log(2n+1)}{2n+1}+\frac{\gamma}{2n+1} \right) \\ &= -\frac{\gamma \pi}{4}-\sum_{n=1}^\infty \frac{\sin\left(\frac{\pi n}{2}\right)\log n}{n} \quad \color{blue}{(3)} \end{aligned} To get the value of the sum, plug $s=\frac{1}{4}$ in equation (1). $$\sum_{n=1}^\infty \frac{\sin\left(\frac{\pi n}{2}\right)\log n}{n} = \pi \log \Gamma\left(\frac{1}{4}\right) - \frac{\gamma \pi}{4} - \frac{\pi \log 2}{2}- \frac{3\pi}{4}\log(\pi)$$ Substituting the above into equation (3) and performing some simplifications gives the desired result.