Integration of $\int^\infty_0\frac{1}{1+\left(x^2+b^2\right)^\mu}dx$

Question

Is it possible to calculate the integral $\int^\infty_0\frac{1}{1+\left(x^2+b^2\right)^\mu}dx$ where $\mu>1$ is a real number? I don't have any clue...

Update: Since a closed-form expression is infeasible. I am looking for an infinite sum expression. I tried the technique of the top answer in Prove $\int_0^{\infty}\! \frac{\mathbb{d}x}{1+x^n}=\frac{\pi}{n \sin\frac{\pi}{n}}$ using real analysis techniques only, however after the variable substitution, there will be an extra $\frac{1}{x}$ term compared with the expression in the linked answer, thus infeasible.

Attempt 2: The integral could be rewritten as $\int_0^{\infty} \int_0^{\infty} e^{-\left(1+(x^2+b^2)^\mu\right) t} d t d x$ where we can expand $e^{-t(x^2+b^2)^\mu} = \sum_n \frac{-t(x^2+b^2)^{n\mu}}{n!}$ and we can perform polynomial expansion again. For $x>b$,$(x^2+b^2)^{\mu n}=\sum_{k=0}^{\infty}\binom{\mu n}{k} x^{\mu n-k} b^k$ ,for $x<b$, $(x^2+b^2)^{\mu n}=\sum_{k=0}^{\infty}\binom{\mu n}{k} b^{\mu n-k} x^k$.

Attempt 3: Perform taylor expansion on $\frac{1}{y+(x^2+b^2)^\mu} = \sum_{n=0}^\infty (-1)^n(x^2+b^2)^{-(n+1)\mu}y^n$ let $y=1$ we have $$ \frac{1}{y+(x^2+b^2)^\mu} = \sum_{n=0}^\infty (-1)^n(x^2+b^2)^{-(n+1)\mu} $$ and $(x^2+b^2)^{-c}$ is integrable.

Answer 1

Using substiting: $t=x^2+b^2$ and Mellin Transform and Inverse Mellin Transform:

$$\int_0^{\infty } \frac{1}{1+\left(x^2+b^2\right)^{\mu }} \, dx=\\\int_{b^2}^{\infty } \frac{1}{\sqrt{-b^2+t} \left(2+2 t^{\mu }\right)} \, dt=\\\mathcal{M}_s^{-1}\left[\int_{b^2}^{\infty } \mathcal{M}_A\left[\frac{1}{\sqrt{-b^2+t} \left(2+A t^{\mu }\right)}\right](s) \, dt\right](A)=\\\mathcal{M}_s^{-1}\left[\int_{b^2}^{\infty } \frac{2^{-1+s} \pi \left(t^{\mu }\right)^{-s} \csc (\pi s)}{\sqrt{-b^2+t}} \, dt\right](A)=\\\mathcal{M}_s^{-1}\left[\frac{2^{-1+s} b^{1-2 s \mu } \pi ^{3/2} \csc (\pi s) \Gamma \left(-\frac{1}{2}+s \mu \right)}{\Gamma (s \mu )}\right](A)=\\\mathcal{M}_s^{-1}\left[\frac{b \left(\frac{1}{2 b^{-2 \mu }}\right)^{-s} \sqrt{\pi } \Gamma (1-s) \Gamma (s) \Gamma \left(-\frac{1}{2}+s \mu \right)}{2 \Gamma (s \mu )}\right](2)=\\\frac{1}{2}| b|\sqrt{\pi }H^{2,1}_{2,2} \left[b^{2 \mu } \Bigg| \begin{matrix} (0,1), & (0,\mu) \\ (-\frac{1}{2},\mu),&(0,1)) \end{matrix}\right]$$ for: $\mu >\frac{1}{2}$ and $b\in \mathbb{R}$.

where: $H$ is FoxH function.

Answer 2

If $b>1$

$$\frac{1}{1+\left(x^2+b^2\right)^\mu}=\sum_{k=0}^\infty \, \frac {(-1)^k }{ \left(x^2+b^2\right)^{(k+1) \mu }}$$ and, if $m>1$ $$J_m=\int_0^\infty \frac {dx}{ \left(x^2+b^2\right)^m}=\frac{\sqrt{\pi }}{2}\,\,\frac{\Gamma\left(m-\frac{1}{2}\right)}{\Gamma(m)\,\,b^{2m-1}}$$

Replacing $m$ by $(k+1)\mu$, the ratio of consecutive terms is $$b^{-2\mu}\,\Bigg(1-\frac 1{2k}+\frac{7 \mu -3}{8 k^2 \mu } +O\left(\frac{1}{k^3}\right) \Bigg)$$ showing a fast convergence as soon as $b>1$.

Trying for $\mu=b=2$, the partial sums (from $k=0$ to $k=p$) are $$\left( \begin{array}{cc} p & \sum_{k=0}^p \\ 0 & 0.0981747704247 \\ 1 & 0.0943398184550 \\ 2 & 0.0945285699972 \\ 3 & 0.0945185285275 \\ 4 & 0.0945190842078 \\ 5 & 0.0945190527140 \\ 6 & 0.0945190545278 \\ 7 & 0.0945190544221 \\ 8 & 0.0945190544283 \\ 9 & 0.0945190544280 \\ 10 & 0.0945190544280 \\ \end{array} \right)$$

Since it is an alternating series, if you want to know the number of terms to be added in order to have $$R_k=\frac{\sqrt{\pi }}{2}\,\frac{\Gamma \left((k+1) \mu -\frac{1}{2}\right)}{\Gamma ((k+1) \mu )}\,b^{1-2 (k+1) \mu } \leq \epsilon$$ using Stirling approximation $$R_k \sim \frac 12\,b^{1-2 (k+1) \mu }\sqrt{\frac{\pi}{k \mu }}$$ leads to $$k\sim\frac{W(t)}{4\mu\log(b) }\qquad \text{where} \qquad t=\frac{\pi\, b^{2-4 \mu }\, \log (b)}{\epsilon ^2}$$ $W(t)$ being Lambert function.

Edit

In private, @Mariusz Iwaniuk kindly provided the series expansion of the integral for small $b$. It write $$\int^\infty_0\frac{dx}{1+\left(x^2+b^2\right)^\mu}=\frac{\pi }{2 \mu }\csc \left(\frac{\pi }{2 \mu }\right)-\frac \pi \mu \sum_{n=1}^\infty \frac{\alpha_n}{\beta_n}\,\csc \left(\frac{(2 n-1)\pi}{2 \mu }\right)\,b^{2n}$$

where, with a shift of $1$, the $\alpha_n$ correspond to sequence $A001790$ and the $\beta_n$ correspond to sequence $A101926$ in $OEIS$.