# Complex Integral

• May 3rd 2010, 04:17 PM
zizou1089
Complex Integral

$J=\int_{-\infty }^{\infty }\frac{x^{2}dx}{1+x^{6}}$

Is it right to say that we use a contour consisting of the real interval [-R,R].

and to write the integral again in terms of f(z)?
• May 3rd 2010, 05:30 PM
Soroban
Hello, zizou1089!

Quote:

$J=\int_{-\infty }^{\infty }\frac{x^2\,dx}{1+x^6}$

We have: . $\int\frac{x^2\,dx}{1 + (x^3)^2}$

Let: . $u \:=\:x^3 \quad\Rightarrow\quad du \:=\:3x^2\,dx \quad\Rightarrow\quad x^2\,du \:=\:\tfrac{1}{3}du$

Substitute: . $\int\frac{\frac{1}{3}\,du}{1+u^2} \;=\;\tfrac{1}{3}\arctan u + C$

Back-substitute: . $\tfrac{1}{3}\arctan(x^3) + C$

Evaluate: . $\tfrac{1}{3}\arctan(x^3)\,\bigg]^{\infty}_{\text{-}\infty}$

. . . . . . $=\;\lim_{b\to\infty} \tfrac{1}{3}\arctan(x^3)\,\bigg]^b_{\text{-}b}$

. . . . . . $= \;\frac{1}{3}\cdot\lim_{b\to\infty}\bigg[\arctan(b) - \arctan(\text{-}b)\bigg]$

. . . . . . $=\;\frac{1}{3}\bigg[\frac{\pi}{2} - \left(\text{-}\frac{\pi}{2}\right)\bigg]$

. . . . . . $=\;\frac{\pi}{3}$