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I need a closed form for the following integral

$$\int^\infty_0\frac{t+b}{(t^2+a)(k^2+(t+b)^2)}dt$$

where $a,b \in \mathbb{C}$ and $k \in \mathbb{N}$. That $t+b$ makes the integral complicated for me.

Any simple approaches for that ?

I tried wolfram |Alpha but it fails. A final solution using CAS is acceptable for me.

I guess a closed form can be obtained using Incomplete Beta function, But I need a simpler one ?

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    call your integrand $f(t,a,b)$. Now integrate $\log(t)f(t,a,b)$ around a keyhole contour in the complex plane. Observe that as $|t|\rightarrow \infty, f(t,a,b)\sim C |t|^{-3}$ so a huge portion of this contour vanishs... – 2017-01-06

1 Answers 1

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Use partial fractions to do the calculation. Let $$ \frac{t+b}{(t^2+a)(k^2+(t+b)^2)}=\frac{At+B}{t^2+a}+\frac{Ct+D}{(t+b)^2+k^2}$$ and then one has $$ (At+B)((t+b)^2+k^2)+(Ct+D)(t^2+a)=t+b.$$ It is not hard to obtain \begin{eqnarray} A&=&\frac{k^2-a-b^2}{a^2+2 a b^2-2 ak^2+b^4+2 b^2 k^2+k^4},\\ B&=&\frac{a b+b^3+b k^2}{a^2+2 a b^2-2 a k^2+b^4+2 b^2 k^2+k^4},\\ C&=&-A,\\ D&=&\frac{a b+b^3-3 b k^2}{a^2+2 a b^2-2 a k^2+b^4+2 b^2 k^2+k^4}. \end{eqnarray} (Suppose $a,b,k>0$.) It is not hard to get \begin{eqnarray} &&\int^\infty_0\frac{t+b}{(t^2+a)(k^2+(t+b)^2)}dt\\ &=&\int^\infty_0\left(\frac{At+B}{t^2+a}+\frac{-At+D}{(t+b)^2+k^2}\right)\\ &=&\frac A2\ln\frac{t^2+a}{k^2+(t+b)^2}+\frac{B}{\sqrt a}\arctan\frac{t}{\sqrt a}+\frac{-bC+D}{k}\arctan\frac{b+t}{k}\bigg|_0^\infty\\ &=&(\frac{B}{\sqrt a}+\frac{-bC+D}{k})\frac{\pi}{2}-\frac{A}{2}\ln\frac{a}{k^2+b^2}-\frac{-bC+D}{k}\arctan\frac{b}{k}. \end{eqnarray}