<s>[INST] $\int_{0}^{\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}$ equals $\frac{\pi}{5050}$ For $a\geq2$,if the value of the definite integral $\int_{0}^{\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}$ equals $\frac{\pi}{5050}$.Find the value of $a$.
Substituting $x-\frac{1}{x}=t$ does not seem to work here,what is the good substitution to make it integrable?Thanks in advance.
 [/INST] Let $$\displaystyle I = \int_{0}^{\infty}\frac{1}{a^2+\left(x-\frac{1}{x}\right)^2}dx = \int_{0}^{\infty}\frac{1}{x^2+\frac{1}{x^2}+(a^2-2)}dx\;,$$ Where $a^2-2 = k \geq 0$
So $$\displaystyle I = \int_{0}^{\infty}\frac{x^2}{x^4+kx^2+1}dx = \frac{1}{2}\int_{0}^{\infty}\frac{(x^2+1)+(x^2-1)}{x^4+kx^2+1}dx$$
So we get $$\displaystyle I = \underbrace{\int_{0}^{\infty}\frac{x^2+1}{x^4+kx^2+1}dx}_{I}+\underbrace{\int_{0}^{\infty}\frac{x^2-1}{x^4+kx^2+1}dx}_{J}$$
Now $$\displaystyle I = \int_{0}^{\infty}\frac{x^2+1}{x^4+kx^2+1}dx = \int_{0}^{\infty}\frac{1+x^{-2}}{x^2+x^{-2}+k} = \int_{0}^{\infty}\frac{(1+x^{-2})}{\left(x-\frac{1}{x}\right)^2+\left(\sqrt{k+2}\right)^2}dx$$
Now Substute $$\displaystyle \left(x-\frac{1}{x}\right) = t\;,$$ Then $\left(1+\frac{1}{x^2}\right)dx = dt$ and Changing Limit, We get
and also Put $a^2-2 = k\Rightarrow a^2=k+2.$
$$\displaystyle J = \int_{-\infty}^{\infty}\frac{1}{t^2+a^2}dt = 2\int_{0}^{\infty}\frac{1}{t^2+a^2}dt = \frac{\pi}{2a}$$
Similarly for $$\displaystyle J = \int_{0}^{\infty}\frac{x^2-1}{x^4+kx^2+1}dx = $$
Put $\displaystyle x= \frac{1}{u}\;,$ Then $\displaystyle dx = -\frac{1}{u^2}du$ and Changing Limit, We get
$$\displaystyle K = -\int_{\infty}^{0}\frac{1-u^2}{u^4+ku^2+1}du = \int_{0}^{\infty}\frac{1-u^2}{u^4+ku^2+1}du = -\int_{0}^{\infty}\frac{u^2-1}{u^4+ku^2+1}du $$
so we get $$\displaystyle K = -\int_{0}^{\infty}\frac{x^2-1}{x^4+kx^2+1}dx=-K $$
above we have used the formulae
$\displaystyle \int_{a}^{b} f(x)dx = -\int_{b}^{a} f(x)dx $ and $\displaystyle \int_{a}^{b} f(t)dt = \int_{a}^{b} f(x)dx  $
So we get $K=0$
So $$\displaystyle I = J+K = \frac{\pi}{2a}+0 = \frac{\pi}{5050}$$
so we get $\displaystyle a= 2525$
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