Prove that integral is independant of its parameter

Question

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.

Answer 1

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 $$ \begin{align} 2A &=\int_0^\infty\frac{(1+x^\alpha)\,\mathrm{d}x}{(1+x^2)(1+x^\alpha)}\\ &=\int_0^\infty\frac{\mathrm{d}x}{(1+x^2)}\\ &=\frac\pi2\tag{3} \end{align} $$

Answer 2

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

$$ \int_{\infty}^1 {\frac {d \left( \frac 1 y\right)} {(1+y^{-2})(1+{y}^{-\alpha})}} = \int_{\infty}^1 \frac{- y^2 y^{\alpha}\frac 1 {y^2}}{(y^2+1)(y^\alpha + 1)}\, dy = \int_1^{\infty}\frac{y^{\alpha}}{(1+y^2)(1+y^\alpha)}\, dy$$

Answer 3

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{\dd x \over \pars{1 + x^{2}}\pars{1 + x^{\alpha}}}}$

\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}

Answer 4

Add the two fractions together and you get

$$J_1+J_2 = \int_1^\infty {\left(\frac{1}{1+x^\alpha}+\frac{x^\alpha}{1+x^\alpha}\right) \frac {dx}{1+x^2}}=\int_1^{\infty}\frac{1+x^{\alpha}}{1+x^{\alpha}}\frac {dx}{1+x^2}=\int_1^{\infty}\frac {dx}{1+x^2}$$