Integral of the day XXXI

Integral involving a partial fraction expansion

Today we show the proof of this integral posted by @integralsbot $$\int_{0}^{\infty} \prod_{k=1}^{\infty} \frac{k^2}{x^2+k^2} dx = \frac{\pi n}{2(2n-1)}$$ The proof relies on the partial fraction decomposition and some properties of the binomial coefficient.

Proof:

We need the following partial fraction decomposition (inspired by this this exercise ): $$\prod_{k=1}^{n} \frac{1}{x+k^2} = 2\sum_{k=1}^{n}\frac{(-1)^{k-1}k^2}{(x+k^2)(n-k)!(n+k)!}$$

Here is the proof:

Since $\deg 1 \lt \deg (x+1)(x+2^2)\cdots (x+n^2)$ there exists a partial fraction decomposition: $$\frac{1}{(x+1)(x+2^2)\cdots(x+n^2)} = \sum_{k=1}^{n} \frac{A_{k}}{x+k^2}$$ Multiplying by $(x+1)(x+2^2)\cdots (x+n^2)$ $$1= \sum_{k=1}^{n} A_{k}\prod_{i=1 \atop i\neq k}^n (x+i^2)$$ Suppose that $x = -m^2$ with $1\leq m \leq n$ Hence $$\begin{align*} 1= & A_{m}\prod_{i=1 \atop i\neq m}^n (-m^2+i^2) \\ =& A_{m} (-m^2+1)(-m^2+2^2)+\cdots(-m^2+(m-1)^2)(-m^2+(m+1)^2)\cdots (-m^2+n^2)\\ =& A_{m} \left(-(m^2-1)\right)\left(-(m^2-2^2)\right)\cdots\left(-(m^2-(m-1)^2)\right)\left((m+1)^2-m^2\right)\cdots\left(n^2-m^2\right)\\ =& A_{m}(-1)^{m-1} (m-1)(m+1)(m-2)(m+2)\cdots\left(m-(m-1)\right)\left(m+(m-1)\right)\left((m+1)+m\right)\left((m+1)-m\right)\cdots (n-m)(n+m)\\ =& A_{m}(-1)^{m-1}\left[(m-1)(m-2)\cdots 2\cdot 1 \cdot 2 \cdots (n-m)\right]\left[(m+1)(m+2)\cdots \left(m+(m-1)\right)\left(m+(m+1)\right)\cdots (n+m)\right]\\ =& A_{m} (-1)^{m-1}\left[(m-1)!(n-m)!\right]\left[(m+1)(m+2)\cdots \left(m+(m-1)\right)\left(m+(m+1)\right)\cdots (n+m)\right]\\ \end{align*}$$ Multiplying and dividing by $\displaystyle \frac{(2m) m!}{(2m) m!}$ $$\begin{align*} 1 = &\frac{A_{m} (-1)^{m-1}\left[(m-1)!(n-m)!\right]\left[m!(m+1)(m+2)\cdots \left(m+(m-1)\right)(2m)\left(m+(m+1)\right)\cdots (n+m)\right]}{(2m)m!}\\ =& \frac{A_{m} (-1)^{m-1}\left[(m-1)!(n-m)!\right]\left[(n+m)!\right]}{(2m)m!} \end{align*}$$ Multiplying and dividing by $m$ $$\begin{align*} 1 = & \frac{A_{m} (-1)^{m-1}m!(n-m)!(n+m)!}{(2m^2)m!}\\ =& \frac{A_{m} (-1)^{m-1}(n-m)!(n+m)!}{2m^2} \end{align*}$$ Therefore $$A_{m} = \frac{2(-1)^{m-1}m^2}{(n-m)!(n+m)!} \quad 1\leq m \leq n$$ Hence $$\prod_{k=1}^{n}\frac{1}{x+k^2} = 2\sum_{k=1}^{n} \frac{(-1)^{n-1}k^2}{(x+k^2)(n-k)!(n+k)!}$$ Back to the integral: $$\begin{align*} I = \int_{0}^{\infty} \prod_{k=1}^{\infty} \frac{k^2}{x^2+k^2} dx = & \left(\prod_{k=1}^{n}k^2 \right) \int_{0}^{\infty} \prod_{k=1}^{\infty} \frac{1}{x^2+k^2} dx\\ =& 2n!^2 \int_{0}^{\infty} \sum_{k=1}^{n} \frac{(-1)^{n-1}k^2}{(x^2+k^2)(n-k)!(n+k)!} dx\\ =& 2n!^2 \sum_{k=1}^{n} \frac{(-1)^{n-1}k^2}{(n-k)!(n+k)!}\int_{0}^{\infty} \frac{1}{x^2+k^2} dx\\ =& 2n!^2 \sum_{k=1}^{n} \frac{(-1)^{n-1}}{(n-k)!(n+k)!}\int_{0}^{\infty} \frac{1}{\left(\frac{x}{k}\right)^2+1} dx\\ =& 2n!^2 \sum_{k=1}^{n} \frac{(-1)^{n-1}k}{(n-k)!(n+k)!}\int_{0}^{\infty} \frac{1}{w^2+1} dx \quad \left(w \mapsto \frac{x}{k}\right)\\ =& 2n!^2 \sum_{k=1}^{n} \frac{(-1)^{n-1}k}{(n-k)!(n+k)!}\left[\arctan(w)\right]_{0}^{\infty}\\ =& \pi n!^2 \sum_{k=1}^{n} \frac{(-1)^{n-1}k}{(n-k)!(n+k)!} \end{align*}$$ where we have used the fact that $\displaystyle \prod_{k=1}^{n} k^2 = \left(\prod_{k=1}^{n} k\right)\left(\prod_{k=1}^{n} k\right) = n!^2$

Finally, we have to prove that $$\pi n!^2 \underbrace{\sum_{k=1}^{n} \frac{(-1)^{n-1}k}{(n-k)!(n+k)!}}_{S} = \frac{\pi n}{2(2n-1)}$$ Note that $$\begin{align*} S =\frac{1}{(2n)!}\sum_{k=1}^{n} \frac{(-1)^{k-1}k(2n)!}{(n-k)!(n+k)!} =& \frac{1}{(2n)!}\sum_{k=1}^{n} \binom{2n}{n+k}(-1)^{k-1}k\\ =& \frac{1}{(2n)!}\sum_{j=n+1}^{2n} \binom{2n}{j}(-1)^{n-j-1}(j-n) \quad (j=n+k)\\ =& \frac{1}{(2n)!}\sum_{j=n+1}^{2n} \binom{2n}{2n-j}(-1)^{n-j-1}(j-n) \quad \left(\binom{v}{v-w} = \binom{v}{w} \right)\\ =& \frac{1}{(2n)!}\sum_{m=n-1}^{m=0} \binom{2n}{m}(-1)^{m-n-1}(n-m) \quad (m=2n-j) \\ =& \frac{(-1)^{n-1}}{(2n)!}\sum_{m=0}^{n-1} \binom{2n}{m}(-1)^{m}(n-m)\\ =& \frac{(-1)^{n-1}}{(2n)!}\left[n\sum_{m=0}^{n-1} \binom{2n}{m}(-1)^{m}-\sum_{m=0}^{n-1} \binom{2n}{m}(-1)^{m}m \right]\\ =& \frac{(-1)^{n-1}}{(2n)!}\left[n\sum_{m=0}^{n-1} \binom{2n}{m}(-1)^{m}-2n\sum_{m=0}^{n-1} \binom{2n-1}{m-1}(-1)^{m} \right]\\ =& \frac{(-1)^{n-1}}{(2n)!}\left[2n\sum_{m=0}^{n-1} \binom{2n-1}{m}(-1)^{m}-n\sum_{m=0}^{n-1} \binom{2n}{m}(-1)^{m} \right] \quad \left(\binom{v+1}{w} = \binom{v}{w} + \binom{v}{w-1} \right)\\ =& \frac{1}{(2n)!}\left[2n\binom{2n-2}{n-1}- n\binom{2n-1}{n-1} \right] \quad \left(\sum_{j=0}^{m} (-1)^j\binom{v}{j} = (-1)^n\binom{v-1}{m}\right)\\ =& \frac{1}{(2n)!}\left[\frac{1}{2n-1}\frac{(2n)!}{(n-1)!(n-1)!}- \frac{1}{2n}\frac{(2n)!}{(n-1)!(n-1)!} \right]\\ =& \frac{n}{2(2n-1)n!n!} \end{align*}$$ Hence $$I =\int_{0}^{\infty} \prod_{k=1}^{\infty} \frac{k^2}{x^2+k^2} dx = \pi n!^2\sum_{k=1}^{n} \frac{(-1)^{k-1}k}{(n-k)!(n+k)!} = \pi n!^2 S = \frac{\pi n}{2(2n-1)}$$ Therefore, we can conclude $$\boxed{\int_{0}^{\infty} \prod_{k=1}^{\infty} \frac{k^2}{x^2+k^2} dx = \frac{\pi n}{2(2n-1)}}$$