A very easy integral

Let $J$ denote the integral. Taking integration by parts, $J = \int_{0}^{\infty}\arctan(1/x^2) \, \mathrm{d}x = \underbrace{ \left[ x \arctan(1/x^2) \right]_{0}^{\infty} }_{=0} + \int_{0}^{\infty} \frac{2x^2}{1+x^4} \, \mathrm{d}x = \int_{-\infty}^{\infty} \frac{x^2}{1+x^4} \, \mathrm{d}x.$ At this point, the last integral can be solved completely by elementary means. However, let us give a quicker solution using Glasser's master theorem . Rewriting the integrand, $J = \int_{-\infty}^{\infty} \frac{1}{x^2 + x^{-2}} \, \mathrm{d}x = \int_{-\infty}^{\infty} \frac{1}{\left( x - x^{-1} \right)^2 + 2} \, \mathrm{d}x$ Now, by the Glasser's master theorem, this equals $J = \int_{-\infty}^{\infty} \frac{1}{u^2 + 2} \, \mathrm{d}u = \frac{\pi}{\sqrt{2}}.$

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Addendum. For the reader interested in a completely elementary solution, here is a way to go: Write $\begin{aligned} J &= \int_{0}^{\infty} \frac{2}{x^2+x^{-2}} \, \mathrm{d}x \\ &= \int_{0}^{\infty} \frac{1 - x^{-2}}{x^2+x^{-2}} \, \mathrm{d}x + \int_{0}^{\infty} \frac{1 + x^{-2}}{x^2+x^{-2}} \, \mathrm{d}x \\ &= \int_{0}^{\infty} \frac{1 - x^{-2}}{(x + x^{-1})^2 - 2} \, \mathrm{d}x + \int_{0}^{\infty} \frac{1 + x^{-2}}{(x - x^{-1})^2 + 2} \, \mathrm{d}x. \end{aligned}$ Now we substitute $v = x + x^{-1}$ for the first integral and $u = x - x^{-1}$ for the second integral. Then $\begin{aligned} J &= \int_{\infty}^{\infty} \frac{1}{v^2 - 2} \, \mathrm{d}v + \int_{-\infty}^{\infty} \frac{1}{u^2 + 2} \, \mathrm{d}u. \end{aligned}$ The limits of the first integral may be puzzling at the first glance, but they simply reflect the fact that, as $x$ moves from $0$ to $\infty$ , $v = v(x)$ descends from $\infty$ to the minimun $v(1) = 2$ and then go back to $\infty$ . So in effect, this is simply the sum of two convergent improper integrals which cancel out each other: ${``}\displaystyle \int_{\infty}^{\infty} \frac{1}{v^2 - 2} \, \mathrm{d}v \, \text{''} \, = \int_{\infty}^{2} \frac{1}{v^2 - 2} \, \mathrm{d}v + \int_{2}^{\infty} \frac{1}{v^2 - 2} \, \mathrm{d}v = 0.$ So we are led to the same last step as in the previous solution, hence obtaining the answer $\frac{\pi}{\sqrt{2}}$ .

Similar solution with @Aaghaz Mahajan's

Using Feynman's integration trick of differentiation under the integral sign as follows:

$\begin{aligned} I(a) & = \int_0^\infty \tan^{-1} \left(\frac a{x^2}\right)\ dx \\ \frac {\partial I(a)}{\partial a} & = \int_0^\infty \frac {x^2}{a^2+x^4} \ dx & \small \color{#3D99F6} \text{Let }u = \frac {x^4}{a^2} \implies a^2du = 4x^3 dx \\ & = \frac 1{4\sqrt a} \int_0^\infty \frac {u^{-\frac 14}}{1+u} \ du & \small \color{#3D99F6} \text{Beta function: }B(m,n) = \int_0^\infty \frac {u^{m-1}}{(1+u)^{m+n}} \ du \\ & = \frac 1{4\sqrt a} B \left(\frac 34, \frac 14 \right) & \small \color{#3D99F6} B(m,n) = \frac {\Gamma (m) \Gamma (n)}{\Gamma (m+n)} \text{, where }\Gamma (\cdot) \text{ denotes gamma function} \\ & = \frac {\color{#3D99F6}\Gamma \left(\frac 34 \right) \Gamma \left(\frac 14 \right)}{4\sqrt a \color{#D61F06} \Gamma (1)} & \small \color{#3D99F6} \text{Legendre duplication: } \sqrt \pi \Gamma (2z) = 2^{2z-1} \Gamma(z) \Gamma \left(z + \frac 12\right) \\ & = \frac {\color{#3D99F6}\sqrt{2\pi}\Gamma \left(\frac 12 \right)}{4\sqrt a \color{#D61F06}(0!)} & \small \color{#3D99F6} \text{with }z = \frac 14 \color{#D61F06} \text{ and by }\Gamma (n) = (n-1)! \\ & = \frac \pi{\sqrt{8a}} & \small \color{#3D99F6} \text{Since }\Gamma \left(\frac 12 \right) = \sqrt \pi \\ \implies I(a) & = \int \frac \pi{\sqrt{8a}} \ da \\ & = \frac \pi{\sqrt 2}\sqrt a + C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ I(0) & = 0 + C = 0 & \small \color{#3D99F6}\implies C = 0 \end{aligned}$

Therefore, $\displaystyle I(1) = \int_0^\infty \tan^{-1} \left(\frac 1{x^2}\right)\ dx = \frac \pi{\sqrt 2}$ , $\implies A+B = 1 + 2 = \boxed 3$ .

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**References:

• Beta function
• Gamma function

Let $\displaystyle J=\int_0^{\infty} \arctan \left(\frac {1}{t^2}\right)dt$ Substituting $\displaystyle \frac {1}{t^2}= u$ the integral changes to $\displaystyle J=\frac 12\int_0^{\infty} u^{-\frac 32}\arctan u du=\frac 12\int_0^{\infty} u^{-\frac 32}\sum_{k=0}^{\infty} \frac {(-1)^k}{2k+1} u^{2k+1} du$

$\displaystyle = \frac 12\int_0^{\infty} u^{-\frac 12}\sum_{k=0}^{\infty} \frac {(-u^2)^k}{k!}\cdot \frac {\Gamma(k+1)}{2k+1} du$

Using substitution $u^2=x$ this changes to $\displaystyle J=\frac 14 \int_0^{\infty} x^{-\frac 34}\sum_{k=0}^{\infty} \frac {\Gamma(k+1)}{2k+1} \frac {(-x)^k}{k!} dx$

Using Ramanujan's Master theorem this integral equals to $\displaystyle J=\frac 14 \Gamma(s)\phi(-s)$ where $s=\frac 14$ and $\displaystyle \phi(k)=\frac {\Gamma(k+1)}{2k+1}$

Substituting the values you get the answer as $\displaystyle J=\frac {\pi}{\sqrt 2}$

Hence $A=1$ and $B=2$

For those who wish to do this without applying RMT, let

$\displaystyle F\left(a\right)=\int_0^{\infty}\arctan\left(\frac{a}{x^2}\right)dx$
Now, differentiating under the integral sign leads us to

$\displaystyle F'\left(a\right)=\int_0^{\infty}\frac{x^2}{x^4+a^2}dx$

Substituting $x^2=a\tan t$ we get,

$\displaystyle F'\left(a\right)=\frac{1}{4\sqrt{a}}\int_0^{\frac{\pi}{2}}2\left(\sin t\right)^{\frac{1}{2}}\cdot\left(\cos t\right)^{-\frac{1}{2}}dt$

This is a simple application of Beta function along with Euler's Reflection Formula and hence, we get,

$\displaystyle F'\left(a\right)=\frac{\pi}{2\sqrt{2a}}$

Now, integrating the expression, and using $F\left(0\right)=0$ we get

$\displaystyle F\left(a\right)=\pi\sqrt{\frac{a}{2}}$