Integrals and Series

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.

Consider the sequence $\{B_r(x)\}_{r=0}^{\infty}$ of polynomials defined using the recursion:

The first few Bernoulli Polynomials are: $$ \begin{aligned} B_0(x) & =1 \\ B_1(x) & =x-\frac{1}{2} \\ B_2(x) & =x^2-x+\frac{1}{6} \\ B_3(x) & =x^3-\frac{3}{2}x^2+\frac{1}{2}x \\ B_4(x) & =x^4-2x^3+x^2-\frac{1}{30} \\ \end{aligned} $$

The numbers $B_n = B_n(0)$ are called the Bernoulli numbers. Integrating the relation $B_r^\prime(x)=rB_r(x)$ between 0 and 1 gives: $$B_r(1)-B_r(0) = \int_0^1 B_r^\prime (x) dx = r\int_0^1 B_{r-1}(x) dx = 0 \quad \forall r\geq 2$$ This motivates us to define the periodic Bernoulli polynomials by $$\tilde{B}_r(x) = B_r(\langle x\rangle), \quad x\in\mathbb{R}, \; r\geq 2$$ where $\langle x\rangle$ denotes the fractional part of $x$. We will now compute the Fourier series of $\tilde{B}_r(x)$, where $r\geq 2$. The $n$-th Fourier coefficient is given by: $$ \begin{aligned} a_n &= \int_0^1 \tilde{B}_r(x) e^{-2\pi i n x} dx= \int_0^1 B_r(x) e^{-2\pi i n x} dx \end{aligned} $$ Let’s first consider the case when $n\neq 0$. Integration by parts, gives us: $$ \begin{aligned} a_n &= -\frac{e^{-2\pi i n x}}{2\pi i n}B_r(x)\Big|_0^1 + \frac{1}{2\pi i n}\int_0^1 B_r^\prime (x) e^{-2\pi i n x} dx \\ &= \frac{1}{2\pi i n}\int_0^1 B_r^\prime (x) e^{-2\pi i n x} dx \\ &= \frac{r}{2\pi i n}\int_0^1 B_{r-1}(x) e^{-2\pi i n x} dx \quad \quad (1) \end{aligned} $$ The repeated use of equation (1) gives: $$ \begin{aligned} a_n &= \frac{r!}{(2\pi i n)^{r-1}} \int_0^1 B_1(x) e^{-2\pi i nx} dx \\ &= \frac{r!}{(2\pi i n)^{r-1}} \int_0^1 \left(x-\frac{1}{2} \right) e^{-2\pi i nx} dx \\ &= \frac{r!}{(2\pi i n)^{r-1}}\int_0^1 x e^{-2\pi i n x} dx \\ &= -\frac{r!}{(2\pi i n)^r} \end{aligned} $$ When $n=0$, we have $a_0 = \int_0^1 B_r(x)dx = 0$. Note that the Fourier series $$ -r! \sum_{\substack{n=-\infty \\ n\neq 0}} \frac{e^{2\pi i n x}}{(2\pi i n)^r}$$ converges absolutely for all $r\geq 2$. Therefore, it converges uniformly to $\tilde{B}_r(x)$ for all $r\geq 2$. This leads to the following bound: $$|\tilde{B}_r(x)| \leq \frac{2r!}{(2\pi)^r}\sum_{n=1}^\infty \frac{1}{n^r} < \frac{4r!}{(2\pi)^r} \;\; \forall r\geq 2\quad\quad (2)$$

Note that the above inequality also remains valid for $r=0$ and $r=1$. Now, let’s consider the generating function: $$F(x,t) = \sum_{n=0}^\infty \frac{\tilde{B}_n(x) t^n}{n!} $$ The inequality (2) implies that: $$ |F(x,t)| < 4 \sum_{n=0}^\infty \left(\frac{t}{2\pi}\right)^n $$ Therefore, the series converges uniformly for all $t\in [0,2\pi]$ and all $x$. We, may, therefore differentiate term by term to obtain: $$ \frac{\partial F(x,t)}{\partial x} = \sum_{n=1}^\infty \frac{\tilde{B}_{n-1}(x)}{(n-1)!}t^n = t F(x,t) $$ Solving the above differential equation, we get $F(x,t) = G(t) e^{xt}$ where $G$ is some arbitrary function of $t$. Next, we integrate $F(x,t)$ between 0 and 1: $$ \begin{aligned} \int_0^1 F(x,t) dx &= G(t) \int_0^1 e^{xt} dx \\ &= G(t) \frac{e^t-1}{t} \end{aligned} $$ On the other hand, note that: $$\begin{aligned} \int_0^1 F(x,t) dx &= \int_0^1 \sum_{n=0}^\infty \frac{\tilde{B}_n(x) t^n}{n!} dx \\ &= 1 + \sum_{n=1}^\infty \frac{t^n}{n!}\int_0^1 B_n(x) dx \\ &= 1 \end{aligned} $$ Therefore, we obtain $G(t) = \frac{t}{e^t – 1}$ and $$\boxed{F(x,t) = \frac{t e^{xt}}{e^t-1}}$$

An interesting property of the Bernoulli numbers is that $B_{2n+1}=0$ for all $n\geq 1$. To see this, consider: $$\frac{t}{e^t -1} + \frac{t}{2}= 1+\sum_{n=2}^\infty \frac{B_n t^n}{n!}$$ Now, on the left hand side we have an even function of $t$. Therefore, the coefficients of the odd powers of $t$ on the right hand side are equal to 0. Using the Fourier series expansion, we can express the even-index Bernoulli numbers in terms of the Riemann zeta function: $$B_{2n} = \frac{2 (-1)^{n-1} (2n)!}{(2\pi)^{2n}} \zeta(2n)$$

The Bernoulli polynomials satisfy the following recursive equation: $$ {B}_n(x) = \sum_{k=0}^n \binom{n}{k} B_{n-k} x^k $$ This can be proved by noting that: $$ \begin{aligned} \sum_{n=0}^\infty \frac{{B}_n(x) t^n}{n!} &= \frac{te^{xt}}{e^t-1} \\ &= e^{xt} \sum_{n=0}^\infty \frac{B_n t^n}{n!} \\ &= \sum_{m=0}^\infty \frac{(xt)^m}{m!} \sum_{n=0}^\infty \frac{B_n t^n}{n!} \\ &= \sum_{m=0}^\infty \sum_{n=0}^\infty \frac{B_n x^m t^{n+m}}{n! m!} \\ &= \sum_{n=0}^\infty \sum_{k=0}^n \frac{B_k t^n x^{n-k}}{k! (n-k)!} \\ &= \sum_{n=0}^\infty \frac{t^n}{n!}\sum_{k=0}^n \binom{n}{k} B_k x^{n-k} \end{aligned} $$ where $x\in [0,1]$. Now, compare the coefficients of $t^n$ to get the desired result. Plugging in $x=1$, gives the identity: $$ \sum_{k=0}^{n-1} \binom{n}{k} B_k = 0 $$

Now, let’s turn our attention to the integral: $$I=\int_0^{\frac{\pi}{2}}\frac{\sin(2nx)}{\sin^{2n+2}(x)}\cdot \frac{1}{e^{2\pi \cot x}-1} dx$$ where $n\in\mathbb{N}$. We will use the following trigonometric identity: $$ \frac{\sin(2nx)}{\sin^{2n}(x)} =(-1)^n \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r}\cot^{2r-1}(x) $$ Substituting the above into the integral, gives: $$ \begin{aligned} I &= (-1)^n \int_0^{\pi\over 2}\left( \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r}\cot^{2r-1}(x)\right)\frac{\csc^2(x)}{e^{2\pi \cot x}-1}dx \\ &= (-1)^n \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r} \int_0^{\pi \over 2}\cot^{2r-1}(x)\frac{\csc^2(x)}{e^{2\pi \cot x}-1}dx \\ &= (-1)^n \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r} \int_0^\infty \frac{t^{2r-1}}{e^{2\pi t}-1}dt \\ &= (-1)^n \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r}\frac{(2r-1)! \zeta(2r)}{(2\pi)^r }\\ &= (-1)^n \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r} (-1)^{r-1} \frac{B_{2r}}{4r}\\ &= \frac{(-1)^{n-1}}{4}\sum_{r=1}^{n}\binom{2n}{2r-1}\frac{B_{2r}}{r} \\ &= \frac{(-1)^{n-1}}{2(2n+1)}\sum_{r=1}^n \binom{2n+1}{2r} B_{2r} \\ &= \frac{(-1)^{n-1}}{2(2n+1)} \left[\sum_{r=0}^{2n} \binom{2n+1}{r} B_r – \binom{2n+1}{0}B_0 – \binom{2n+1}{1} B_1\right] \\ &= \frac{(-1)^{n-1}}{2(2n+1)} \left[-\binom{2n+1}{0}B_0 – \binom{2n+1}{1} B_1\right] \\ &= \frac{(-1)^{n-1}}{4}\cdot \frac{2n-1}{2n+1} \end{aligned} $$