Evaluate $\int_0^{\infty} \frac{ dx }{(x^4+c)(x^2+1) } $

Question

Consider $$\int_0^{\infty} \dfrac{ dx }{(x^4+c)(x^2+1) } $$

where $c \in \mathbb{R}$. Now, it is possible to evaluate this using complex analysis, but I was wondering if there is way to compute this integral without complex analysis.

I was thinking on partial fractions, but there is no easy form to decompose it too.

Here is a way I was thinking: Write $1 = \dfrac{c+x^4 - x^4 }{c}$ to obtain

$$ \frac{1}{c} \int_0^{\infty} \frac{dx}{x^2+1} - \dfrac{1}{c} \int_0^{\infty} \dfrac{x^4}{(x^4+c)(x^2+1)} = \dfrac{ \pi }{2c} - \dfrac{1}{c} \int_0^{\infty} \dfrac{x^4}{(x^4+c)(x^2+1)}$$

I dont know if this is "progress", what do you think?

Answer 1

Assume $c>0$ for convergence.

$$I = \int_0^{\infty} \dfrac{ dx }{(x^4+c)(x^2+1) } = \frac1{c+1}\int_0^{\infty}\left( \frac{ 1}{x^2+1}- \frac{ x^2-1}{x^4+c }\right) dx \\ = \frac{\pi}{2(c+1)} - \frac{1}{c+1}\int_0^{\infty}\frac{ x^2-1}{x^4+c } dx$$

where

\begin{align} \int_0^{\infty}\frac{ x^2-1}{x^4+c } dx & =\frac{\sqrt c+1}{2\sqrt c} \int_0^{\infty}\frac{ 1-\frac{\sqrt c}{x^2}}{x^2+\frac c{x^2} } dx + \frac{\sqrt c-1}{2\sqrt c} \int_0^{\infty}\frac{ 1+\frac{\sqrt c}{x^2}}{x^2+\frac c{x^2} } dx \\ & =\frac{\sqrt c+1}{2\sqrt c} \int_0^{\infty}\frac{d(x+\frac{\sqrt c}{x})}{(x+\frac {\sqrt c}{x} )^2-2\sqrt c} + \frac{\sqrt c-1}{2\sqrt c} \int_0^{\infty}\frac{d(x-\frac{\sqrt c}{x})}{(x-\frac {\sqrt c}{x} )^2+2\sqrt c} \\ & =0+ \frac{(\sqrt c-1)\pi}{(2\sqrt c)^{3/2}} \end{align}

Thus,

$$I = \frac{\pi}{2(c+1)} - \frac{1}{c+1}\frac{(\sqrt c-1)\pi}{(2\sqrt c)^{3/2}} =\frac{\pi}{2(c+1)} \left( 1- \frac{\sqrt c-1}{\sqrt2 c^{3/4}} \right)$$

Answer 2

$$\dfrac{ 1 }{(x^4+c)(x^2+1) }=\frac{1}{\left(x^2+1\right) \left(x^2-i \sqrt{c}\right) \left(x^2+i \sqrt{c}\right)}$$ Using partial fractions, it is $$-\frac{1}{2 \left(\sqrt{c}-i\right) \sqrt{c} \left(x^2-i \sqrt{c}\right)}-\frac{1}{2 \left(\sqrt{c}+i\right) \sqrt{c} \left(x^2+i \sqrt{c}\right)}-\frac{i}{2 \left(\sqrt{c}-i\right) \left(\sqrt{c}+i\right) (x-i)}+\frac{i}{2 \left(\sqrt{c}-i\right) \left(\sqrt{c}+i\right) (x+i)}$$ which is not so bad.