bookmark_borderRamanujan’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.

bookmark_borderFourier 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.

bookmark_borderBernoulli numbers and a related integral

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

\begin{aligned} B_0(x) &= 1 \\ B_r^\prime(x) &= r B_{r-1}(x) \quad \forall r\geq 1 \\ \int_0^1 B_r(x) dx &= 0 \quad \forall r\geq 1 \end{aligned}

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}