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.
You may write $$ \int_{0}^{\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}=\frac12\int_{-\infty}^{\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2} $$ then use the fact that, for any integrable function $f$, we have
$$ \int_{-\infty}^{+\infty}f\left(x-\frac{s}{x}\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x, \quad s>0. \tag1 $$
Applying it to $f(x)=\dfrac1{a^2+x^2}$, $a\geq2$, $s=1$, you get
$$ \int_{0}^{\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}=\frac12\int_{-\infty}^{+\infty} \dfrac1{a^2+x^2} \mathrm{d}x=\frac{\pi}{2a}. \tag2 $$
Thus
$$ \frac{\pi}{2a}=\frac{\pi}{5050}\tag3 $$ and it is easy to obtain $a$.
A proof of $(1)$ may be found here.
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 $\displaystyle a= 2525$
The substitution $t=1/x$ works. Then, $dx=-\frac{1}{t^2}dt$ and
$$\begin{align} I&=\int_0^{\infty}\frac{1}{a^2+\left(x-\frac1x\right)^2}\,dx\\\\ &=\int_0^\infty \frac{1}{a^2+\left(x-\frac1x\right)^2}\frac{1}{x^2}\,dx\\\\ &=\frac12\int_0^\infty \frac{1+\frac{1}{x^2}}{a^2+\left(x-\frac1x\right)^2}\,dx\\\\ &=\frac12\int_0^\infty \frac{1}{a^2+\left(x-\frac1x\right)^2}\,d\left(x-\frac1x\right)\\\\ &=\left.\frac1{2a} \arctan\left(\frac{x-\frac1x}{a}\right)\right|_{0}^{\infty}\\\\ &\bbox[5px,border:2px solid #C0A000]{=\frac{\pi}{2a}\text{sgn}(a)} \end{align}$$
If $a>2$, then setting $\frac{\pi}{2a}=\frac{\pi}{5050}$, we obtain
$$\bbox[5px,border:2px solid #C0A000]{a=2525}$$
It works, indeed. If we set $x=\frac{1}{2}\left(y+\sqrt{4+y^2}\right)$ we get: $$ I = \int_{0}^{+\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}=\frac{1}{2}\int_{-\infty}^{+\infty}\frac{dy}{y^2+a^2}\left(1+\frac{y}{\sqrt{y^2+4}}\right) $$ hence: $$ I = \frac{1}{2}\int_{-\infty}^{+\infty}\frac{dy}{y^2+a^2} = \frac{\pi}{2a}$$ since the odd part of the previous integrand function clearly vanishes by symmetry.
Note \begin{eqnarray} 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\\ &=&\int_{0}^{\infty}\frac{x^2}{x^4+kx^2+1}dx. \end{eqnarray} Here $k=a^2-2$. Suppose $a>2$. Letting $x^4+kx^2+1=(x^2+r_1^2)(x^2+r_2^2)$ where $r_1,r_2>0$, we have \begin{eqnarray} I&=&\frac1{r_1^2-r_2^2}\left(r_1^2\int_{0}^{\infty}\frac{1}{x^2+r_1^2}-r_2^2\int_{0}^{\infty}\frac{1}{x^2+r_2^2}\right)\\ &=&\frac1{r_1^2-r_2^2}\frac{\pi(r_1-r_2)}{2}\\ &=&\frac{\pi}{2}\frac1{r_1+r_2}. \end{eqnarray} Noting $r_1^2+r_2^2=a^2-2$ and $r_1^2r_2^2=1$, we have $r_1+r_2=\sqrt{r_1^2+r_2^2+2r_1r_2}=a$. So $$ I=\frac{\pi}{2a}.$$ Letting $\frac{\pi}{2a}=\frac{\pi}{5050}$ gives $a=2525$.
