Closed form: $\displaystyle\int_0^\infty \! \prod_{i=1}^n\frac{1}{a_i^2 + x^2} \, dx$

Question

Can you find the closed form for the following integral

$$ \int_0^\infty \prod_{i=1}^n \frac{1}{a_i^2 + x^2} \, dx = \int_0^\infty \! \frac{dx}{(a_1^2 + x^2) (a_2^2 + x^2) \cdots (a_n^2 + x^2)} $$

where $a_i \in \mathbb{R}$ for any $i\in\mathbb{Z}_{\gt0}$?

For example,

$$ \begin{align} &\int_0^\infty \! \frac{dx}{a^2 + x^2} = \frac\pi2 \frac1a \\[3pt] &\int_0^\infty \! \frac{dx}{(a^2 + x^2)(b^2 + x^2)} = \frac\pi2 \frac1{ab \, (a+b)} \\[3pt] &\int_0^\infty \! \frac{dx}{(a^2 + x^2)(b^2 + x^2)(c^2 + x^2)} = \frac\pi2 \frac{a+b+c}{abc \, (a+b)(b+c)(c+a)} \\[3pt] \end{align} $$

Answer 1

Partial fractions was suggested. If the $a_i$ are all different, then there exist numerators $b_i$ with $$ \prod_{i=1}^n\frac{1}{a_i^2 + x^2}= \sum_{i=1}^n\frac{b_i}{a_i^2+x^2} $$ so that $$ \int_0^\infty \! \prod_{i=1}^n\frac{1}{a_i^2 + x^2} \, dx = \left[\sum_{i=1}^n \frac{b_i}{a_i}\arctan\frac{x}{a_i} \;\right]_{x=0}^\infty = \sum_{i=1}^n \frac{\pi b_i}{2 a_i} $$

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What are the numerators $b_i$? That is a linear algebra problem. $$ b_i = \prod_{j \ne i} \frac{1}{a_j^2-a_i^2} $$

Answer 2

if all $a_k$ was different then since $a_k ^2 +x^2 = 0$ when $a_k^2 =-x^2 $ wee can use partial fractions by cover up method which is $\frac{1}{(y-a)(y-b)} =\frac{1}{a-b}\frac{1}{y-a} +\frac{1}{b-a} \frac{1}{y-b}$ and by induction we can prove that $\prod\limits_{k=1}^n \frac{1}{a_k ^2 +x^2} =\sum\limits _{k=1}^n\left(\frac{1}{a_k ^2 +x^2} \prod\limits_{i=1 | i \neq k}^n \frac{1}{a_i ^2 +a_k^2 }\right)$

$$\int \prod\limits_{k=1}^n \frac{1}{a_k ^2 +x^2}dx = \int \sum\limits _{k=1}^n\left(\frac{1}{a_k ^2 +x^2} \prod\limits_{i=1 | i \neq k}^n \frac{1}{-a_i ^2 +a_k^2 }\right)dx= \sum\limits _{k=1}^n\left(\frac{\pi}{2a_k} \prod\limits_{i=1 | i \neq k}^n \frac{1}{- a_i ^2 +a_k^2 }\right) $$

if only one value of $a_k$ is repeated more than one time (say $r\leq n$ times ) this would be really painful to write a closed form see A treatise on integral calculus page 147 to see how to evaluate this

the way to evaluate this by partial fraction involve Taylor series or long division which is very difficult to calculate for very specific $\{ a_k\}$ so it would be very hard to write it in general form and don't forget this is with only one value $a_k$ is repeated! if more than one value is repeated this would be much more difficult.