The Contour Integration approach to Infinite Series

Posted on July 14, 2020 by Shobhit Bhatnagar

In this post, we will evaluate the series $\sum_{n=0}^\infty \frac{\cot\left(\frac{2n+1}{2}\pi\sqrt{2} \right)}{(2n+1)^3}$ using contour integration. Let’s define $f:\mathbb{C}\to \mathbb{C}$ as $$ f(z) = \frac{\pi \tan(\pi z)\tan(\pi z \theta)}{z^3} $$ where the parameter $\theta$ is a positive irrational number. Let $C_N$ denote the positively oriented square with vertices $(N+1)(1+i),\; (N+1)(-1+i),\; (N+1)(-1-i)$ and $(N+1)(1-i)$. With some effort, one can show that: $$ \begin{aligned} \left|\int_{C_N} f(z) dz\right| &\leq \frac{4\pi}{(N+1)^2} \left(|\tan((N+1)\pi \theta)| + \frac{1}{|\tanh((N+1)\pi) \tanh((N+1)\pi \theta)|}\right) \end{aligned} $$ By Weyl’s equidistribution theorem, the sequence $\{(N+1)\theta \}_{N=1}^\infty $ is equidistributed modulo 1. Therefore, we can choose a subsequence $\{(N_k+1) \theta\}_{k=1}^\infty $ such that $|\tan(\pi (N_k+1) \theta)|$ remains bounded. It follows that: $$ \lim_{k\to \infty}\int_{C_{N_k}} f(z)\; dz = 0 $$ On the other hand, Residue theorem gives us: $$ \frac{1}{2\pi i}\int_{C_{N_k}} f(z) dz = \substack{\displaystyle \text{Res} \\ z=0}f(z) + \sum_{i=-{N_k}}^{N_k} \substack{\displaystyle \text{Res} \\ z=\frac{2i+1}{2}}f(z) + \sum_{|j+\frac{1}{2}|\leq \theta (N_k+1)} \substack{\displaystyle \text{Res} \\ z=\frac{2j+1}{2\theta}}f(z) $$

This means that the sum of residues of $f(z)$ at it’s poles is equal to zero. A simple calculation shows that: $$ \begin{aligned} \substack{\displaystyle \text{Res} \\ z=0}f(z) &= \pi^3 \theta \\ \substack{\displaystyle \text{Res} \\ z=\frac{2n+1}{2}}f(z) &= -8\frac{\tan\left(\frac{\pi \theta}{2}(2n+1) \right)}{(2n+1)^3}, \quad n\in \{0,1,2,\cdots\}\\ \substack{\displaystyle \text{Res} \\ z=\frac{2n+1}{2\theta}}f(z) &= -8\theta^2 \frac{\tan\left(\frac{\pi}{2\theta}(2n+1) \right)}{(2n+1)^3} , \quad n\in \{0,1,2,\cdots\} \end{aligned} $$ Finally, putting everything together gives us the relation: $$ \sum_{n=0}^\infty \frac{\tan\left(\frac{\pi \theta}{2}(2n+1) \right)}{(2n+1)^3} + \theta^2 \sum_{n=0}^\infty \frac{\tan\left(\frac{\pi}{2\theta}(2n+1) \right)}{(2n+1)^3} = \frac{\pi^3 \theta}{16} $$ The final result is obtained by substituting $\theta = \sqrt{2}+1$ in the above equation. $$\boxed{\sum_{n=0}^\infty \frac{\cot\left(\frac{2n+1}{2}\pi\sqrt{2} \right)}{(2n+1)^3} = -\frac{\pi^3}{32\sqrt{2}}} $$

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An alternating Euler Sum

Posted on July 5, 2020 by Shobhit Bhatnagar

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 $0 \leq x < 1$. We don’t need this 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) Similar to 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 to section 7.6 of “Polylogarithms and Associated Functions” by Leonard Lewin). Luckily for us, there is a much simpler was to evaluate this integral between the limits $0$ and $\frac{1}{2}$.

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 easily verify that the end result is: $$I = -\frac{\pi^4}{180}$$ Next, 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: $$ \boxed{ \sum_{n=1}^\infty \frac{H_n^{(2)}}{n^2 2^n} = \frac{\pi^4}{1440}-\frac{\pi^2}{24}\log^2(2) + \frac{\log^4(2)}{24}+ \frac{1}{4}\log(2)\zeta(3) + \text{Li}_4\left(\frac{1}{2}\right) } $$

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Ramanujan’s Master Theorem: Part I

Posted on July 2, 2020 by Shobhit Bhatnagar

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.

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