Prove that integral is independant of its parameter We are given intergral $\int_0^\infty {\frac {dx} {(1+x^2)(1+x^\alpha)}}$ and task is to prove that it's independent of $\alpha$.
Task was too complicated for me, so I had to stick with solution, recommended in the textbook. It goes on like that:
$\int_0^\infty {\frac {dx} {(1+x^2)(1+x^\alpha)}} = \int_0^1 {\frac {dx} {(1+x^2)(1+x^\alpha)}} + \int_1^\infty {\frac {dx} {(1+x^2)(1+x^\alpha)}} = J_1 + J_2$
In $J_1$ we do replacement $y = \frac 1 x$
$J_1 = \int_1^\infty \frac {dy} {-(1+y^2)(1+y^{-\alpha})}$  
Than textbook says that next step is
$J_1+J_2 = \int_1^\infty {(\frac{1}{1+x^\alpha}+\frac{x^\alpha}{1+x^\alpha})  \frac {dx}{1+x^2}}$
From where it's obvious that inital statement is true. Problem is, I completely fail to understand how that summation of $J_1+J_2$ was done.
Edit
To be more explicit, I can't get this: $\int_1^\infty{\frac {dy}{-(1+y^2)(1+y^{-\alpha})}} + \int_1^\infty {\frac {dx} {(1+x^2)(1+x^\alpha)}} = \int_1^\infty {(\frac{1}{1+x^\alpha}+\frac{x^\alpha}{1+x^\alpha})  \frac {dx}{1+x^2}}$  
So I would appreciate some explanations and pointers.
 A: There is no need to break the integral into pieces. Simply use the substitution $t=1/x$:
$$
\begin{align}
A
&=\int_0^\infty\frac{\mathrm{d}x}{(1+x^2)(1+x^\alpha)}\tag{1}\\
&=\int_0^\infty\frac{t^\alpha\,\mathrm{d}t}{(1+t^2)(1+t^\alpha)}\tag{2}
\end{align}
$$
Add $(1)$ and $(2)$ to get
$$
\begin{align}
2A
&=\int_0^\infty\frac{(1+x^\alpha)\,\mathrm{d}x}{(1+x^2)(1+x^\alpha)}\\
&=\int_0^\infty\frac{\mathrm{d}x}{(1+x^2)}\\
&=\frac\pi2\tag{3}
\end{align}
$$
A: Your original integral was split into two parts, $J_1 + J_2$ which you need to show that it is independent of parameter $\alpha$.
By some transformation your integral parts $J_1+J_2$ changes into
$$\int_\infty^{1}{\frac {dy}{-(1+y^2)(1+y^{-\alpha})}} + \int_1^\infty {\frac {dx} {(1+x^2)(1+x^\alpha)}} = \int_1^\infty {\left(\frac{1}{1+x^\alpha}+\frac{x^\alpha}{1+x^\alpha}\right)  \frac {dx}{1+x^2}} \\ 
= \int_1^{\infty} \frac{1}{1+x^2}\, dx$$
The final term is independent of parameter $a$ which you needed to show.
$$ \int_0^1 {\frac {dx} {(1+x^2)(1+x^\alpha)}}$$
Let $y = \frac 1 x $, $x\to 1 \implies y \to 1$ and $x\to 0\implies y\to \infty$, And the integral converts into
$$ \int_{\infty}^1 {\frac {d \left( \frac 1 y\right)} {(1+y^{-2})(1+{y}^{-\alpha})}} = \int_{\infty}^1 \frac{- y^2 y^{\alpha}\frac 1 {y^2}}{(y^2+1)(y^\alpha + 1)}\, dy = \int_1^{\infty}\frac{y^{\alpha}}{(1+y^2)(1+y^\alpha)}\, dy$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{\infty}{\dd x \over \pars{1 + x^{2}}\pars{1 + x^{\alpha}}}}$

\begin{align}
&\color{#00f}{\large%
\int_{0}^{\infty}{\dd x \over \pars{1 + x^{2}}\pars{1 + x^{\alpha}}}}
=\int_{0}^{\pi/2}{\dd\theta \over 1 + \tan^{2\alpha}\pars{\theta}}
\\[3mm]&=\half\bracks{%
\int_{0}^{\pi/2}{\dd\theta \over 1 + \tan^{2\alpha}\pars{\theta}}
+
\int_{0}^{\pi/2}{\dd\theta \over 1 + \tan^{2\alpha}\pars{\pi/2 - \theta}}}
\\[3mm]&=\half\bracks{%
\int_{0}^{\pi/2}{\dd\theta \over 1 + \tan^{2\alpha}\pars{\theta}}
+
\int_{0}^{\pi/2}{\tan^{2\alpha}\pars{\theta} \over 1 + \tan^{2\alpha}\pars{\theta}}\,\dd\theta}=\half\int_{0}^{\pi/2}\dd\theta
=\color{#00f}{\large{\pi \over 4}}
\end{align}

A: Add the two fractions together and you get
$$J_1+J_2 = \int_1^\infty {\left(\frac{1}{1+x^\alpha}+\frac{x^\alpha}{1+x^\alpha}\right)  \frac {dx}{1+x^2}}=\int_1^{\infty}\frac{1+x^{\alpha}}{1+x^{\alpha}}\frac {dx}{1+x^2}=\int_1^{\infty}\frac {dx}{1+x^2}$$
