Let's Integrate It 2!

Let $x=\frac{1}{x}$ $\text{I}=\int_{\infty}^{0} \frac{\frac{1}{x^2}}{1+\frac{1}{x^4}} \left(-\frac{1}{x^2}\right) \text{d}x=\int_{0}^{\infty} \frac{1}{1+x^4} \text{d}x$ Add this to original integral $2\text{I}=\int_{0}^{\infty} \frac{1+x^2}{1+x^4} \text{d}x$ Let $x=x-\frac{1}{x}$ $2\text{I}=\int_{-\infty}^{\infty} \frac{1}{2+x^2} \text{d}x=\frac{\sqrt{2}}{2} \left( \arctan \left(\frac{x}{\sqrt{2}}\right)\right)_{-\infty}^{\infty}=\frac{\sqrt{2}}{2} \pi \ \ \ \ \Longrightarrow \ \ \ \ \text{I}=\frac{\pi}{2 \sqrt{2}}$

Let $C$ be a large semicircle above the real axis in the complex plane. Let $\zeta=e^{\pi i/4}$ . Since $\displaystyle\frac{z^2}{1+z^4}=\frac{z^2}{(z-\zeta)(z-\zeta^3)(z-\zeta^5)(z-\zeta^7)}$ is analytic within $C$ with poles at $\zeta, \zeta^3$ , by Cauchy's integration formula

$\displaystyle \int_C\frac{z^2}{1+z^4}dz=2\pi i\left(\frac{\zeta^2}{(\zeta-\zeta^3)(\zeta-\zeta^5)(\zeta-\zeta^7)}+\frac{\zeta^6}{(\zeta^3-\zeta)(\zeta^3-\zeta^5)(\zeta^3-\zeta^7)}\right)$

$\displaystyle=2\pi i\left(\frac{i}{\sqrt{2}2\zeta\sqrt{2}i}+\frac{-i}{-\sqrt{2}\sqrt{2}i2\zeta^3}\right)=\frac{\pi i}{2}(e^{7\pi i/4}+e^{5\pi i}{4})=\frac{\pi}{\sqrt{2}}$ .

Since $\frac{z^2}{1+z^4}|z|\to 0$ as $|z|$ grows large, as the radius of $C$ grows large the arc does not contribute to the integral. In the limit, we find

$\displaystyle\int_{-\infty}^{\infty}\frac{x^2}{1+x^4}dx=\frac{\pi}{\sqrt{2}}$ .

Therefore,

$\displaystyle\int_{0}^{\infty}\frac{x^2}{1+x^4}dx=\frac{1}{2}\int_{-\infty}^{\infty}\frac{x^2}{1+x^4}dx=\frac{\pi}{2\sqrt{2}}\approx\boxed{1.110}$ .

First of all, let's split the integral into two parts:

$I = \int_{0}^{\infty} \frac{x^{2}}{1+x^{4}} \mathrm{d}x = \int_{0}^{1} \frac{x^{2}}{1+x^{4}} \mathrm{d}x + \int_{1}^{\infty} \frac{x^{2}}{1+x^{4}} \mathrm{d}x$

Let's call the first integral $I_{1}$ and the second $I_{2}$ . Then

$I_{1} = \int_{0}^{1} x^{2} \sum_{n=0}^{\infty} (-1)^{n} x^{4n}\mathrm{d}x = \sum_{n=0}^{\infty} (-1)^{n} \int_{0}^{1}x^{4n+2} \mathrm{d}x$

And

$I_{2} = \int_{1}^{\infty} \frac{1}{x^{2}} \sum_{n=0}^{\infty} (-1)^{n} \frac{1}{x^{4n}} \mathrm{d}x = \sum_{n=0}^{\infty} (-1)^{n} \int_{1}^{\infty} x^{-(4n+2)} \mathrm{d}x$

These integrals are easy to solve, and

$I_{1} = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{4n+3} \qquad;\qquad I_{2} = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{4n+1}$

The result, is, therefore

$I = I_{1} + I_{2} = \sum_{n=0}^{\infty} (-1)^{n} \left \{ \frac{1}{4n+3} + \frac{1}{4n+1} \right \} = \frac{\pi}{2\sqrt{2}}$

Substitute $x=t^{\frac 14}$ we get $I=\frac 14 \int_0^{\infty}\frac {t^{\frac {-1}{4}}}{1+t} dt=\frac 14 B\left(\frac 14,\frac 34\right)=\frac {\pi}{2\sqrt 2}$

Can't we add (1+1\x^(2))+(1-1\x^(2)) in numerator after multiplying and dividing the whole expression by x^(2)...that was easier fr me!

A long winded, but fairly pedestrian, method.

Letting $x = \frac{u}{\sqrt{2}}$ , the integrand transforms to

$\frac{u^{2}\sqrt{2}}{u^{4} + 4}$

Noting that

$u^{4} + 4 = u^4 + 4u^2 + 4 - 4u^2 = (u+2)^{2} - (2u)^{2} = (u^2+2u+2)(u^2-2u+2)$

we can express the integrand, using partial fractions, as \(

\frac{\sqrt{2}}{4} * ( \frac{u}{u^2-2u+2) - \frac{u}{u^2+2u+2}) \)

Using standard integration techniques, this integrates to

$\frac{\sqrt{2}}{4} * (0.5ln1 + \pi)$

and the answer follows.