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\begin{align*} \frac{1}{(x^2+a^2)\cdots(x^2+(a+n)^2)} &= \frac{2\Gamma(2a)}{\Gamma(n)\Gamma(2a+n)}\left(\frac{a}{x^2+a^2}-\frac{2a}{1!}\frac{n-1}{n+2a}\frac{a+1}{x^2+(a+1)^2}\right. \\ & \qquad + \left.\frac{2a(2a+1)}{2!}\frac{(n-1)(n-2)}{(n+2a)(n+2a+1)}\frac{a+2}{x^2+(a+2)^2}-\cdots\right). \end{align*} The preceding was by Ramanujan, appearing in one of his notebooks. How does one prove this?

Especially interesting is motiving the proof: given only the complete fraction on the left, is there a method that makes the right side almost immediately obvious? (Basically, it would be nice if the answers imagined the RHS didn't exist in the above equation).

Meow
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2 Answers2

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Since both the left- and right-hand sides are in terms of $x^2$ we can change variables to $X = x^2$. This is a special case of the question of finding the partial-fraction expansion $$ \frac1{(X+A_1)(X+A_2)\cdots(X+A_n)} = \sum_{i=1}^n \frac{C_i}{X+A_i} $$ for any distinct $A_1,A_2,\ldots,A_n$. The easy way to find $C_i$ is to multiply both sides by $X+A_i$ and then to evaluate at $X = -A_i$. On the right side this isolates $C_i$. On the left side we get the product over $j \neq i$ of $1/(A_j-A_i)$. So $C_i$ must equal this product.

In the present case, each $A_i$ is $(a+i)^2$, so $A_j-A_i = (a+j)^2 - (a+i)^2$, which factors further as $(j-i)(2a+j+i)$. The product of this over $1 \leq j \leq n$ excluding $j=i$ can then be expressed in various ways in terms of factorials and Gamma functions, one of which yields Ramanujan's choice.

Noam D. Elkies
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First of all there is a error in the expression on the left hand side. The correct right hand side is: \begin{equation} lhs=\frac{1}{(x^2+a^2)\dot \cdots \dot (x^2+(a+n-1)^2)} \end{equation} The left hand side is correct. Now from partial fraction decomposition the right hand side clearly equals: \begin{equation} rhs = \sum\limits_{k=0}^{n-1} \frac{1}{x^2+(a+k)^2} \cdot C_k \end{equation} where: \begin{eqnarray} C_k&=&\frac{1}{\prod\limits_{j=0,j\neq k}^{n-1} (-(a+k)^2 + (a+j)^2)}=\frac{1}{\prod\limits_{j=0,j\neq k}^{n-1} (2 a+k+j)(j-k)} \\ &=&\frac{(-1)^{k}}{(2a+k)^{(k)} (k)! (2a+2 k+1)^{(n-1-k)}(n-1-k)!}\\ &=& \frac{1}{\Gamma(n)} \cdot \binom{n-1}{k} \cdot (-1)^k \cdot \frac{\Gamma(2a+k)}{\Gamma(2 a+n+k)} (2a+2 k) \\ &=& \frac{2 \Gamma(2a)}{\Gamma(n) \Gamma(2 a+n)} \cdot \binom{n-1}{k} (-1)^k \cdot \frac{(2 a)^{(k)}}{(2a+n)^{(k)}} \cdot (a+k) \end{eqnarray} which is exactly what we have on the right hand side. Here of course $k=0,\cdots,n-1$.

Przemo
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  • I don't understand how we have$$C_k=\frac 1{\prod\limits_{j=0}^{n-1}{(a+j)^2-(a+k)^2}}$$And how we have$$\frac 1{\prod\limits_{j=0}^{n-1}(2a+k+j)(j-k)}=\frac {(-1)^k}{(2a+k)^{(k)}k!(2a+2k+1)^{(n-k-1)}(n-k-1)!}$$ – Crescendo Jul 12 '17 at 19:24
  • The first one comes from the generic formula for partial fraction decomposition \begin{equation} \frac{1}{\prod\limits_{j=1}^n(x-b_j)} = \sum\limits_{i=1}^n \frac{1}{x-b_i} \cdot \prod\limits_{i\ne j} \frac{1}{b_i-b_j} \end{equation} and the second one comes from splitting the product in the denominator into firstly a product over $j=0,\cdot, k-1$ and the over $j=k+1,\cdots,n$. The first product produces the the first two terms on the left in the denominator and the second product the second two terms. – Przemo Jul 13 '17 at 09:41