Show that $\int_{-\infty}^{\infty}\frac{x^2dx}{(x^2+1)^2(x^2+2x+2)}=\frac{7\pi}{50}$

Question

Show that

$$\int_{-\infty}^{\infty}\frac{x^2dx}{(x^2+1)^2(x^2+2x+2)}=\frac{7\pi}{50} $$

So I figured since it's an improper integral I should change the limits

$$\lim_{m_1\to-\infty}\int_{m_1}^{0}\frac{x^2dx}{(x^2+1)^2(x^2+2x+2)}+ \lim_{m_2\to\infty}\int_{0}^{m_2}\frac{x^2dx}{(x^2+1)^2(x^2+2x+2)}$$

I'm however not sure how to evaluate this. Any help would be great - thanks.

Answer 1

Hint

Use partial fractions to write out the integral in the form: $$\frac{A x+B}{x^2+1}+\frac{C x+D}{\left(x^2+1\right)^2}+\frac{E+F x}{x^2+2 x+2}$$ You'll find that $a=-6/25, c = 2/5$, etc. Now look at at what remains, and use the fact that: $$\int \frac{1}{x^2+1}dx=\arctan x$$ And substitute $x\rightarrow x+1$ in the last integral to simplify it.

Answer 2

For more manageable partial decomposition, first simplify the integrand with $$x^4+4=(x^2+2x+2)(x^2-2x+2)$$

\begin{align} &\int_{-\infty}^{\infty}\frac{x^2}{(x^2+1)^2(x^2+2x+2)}dx\\ = &\int_{-\infty}^{\infty}\frac{x^4+2x^2}{(x^2+1)^2(x^4+4)}dx\\ = &\ \frac1{25}\int_{-\infty}^{\infty}\frac{2x^2+28}{x^4+4}-\frac{2x^2+7}{(x^2+1)^2} \ dx\\ =& \ \frac1{25}\left( 2+14 -2- 7\right)\frac\pi2=\frac{7\pi}{50} \end{align} where $ \int_{-\infty}^{\infty}\frac1{(x^2+1)^2}dx =\int_{-\infty}^{\infty}\frac{x^2}{(x^2+1)^2}dx = \int_{-\infty}^{\infty}\frac{x^2}{x^4+4}dx =\int_{-\infty}^{\infty}\frac2{x^4+4}dx =\frac\pi2$.

Answer 3

Via complex analysis.

This function has poles at $\pm i$ and $-1 \pm i$

Use the contour of the real line and the semi-circle in the upper half-plane.

The integral along the semi-circular contour will go to 0 as the radius goes to infinity because the degree of the denominator is 6 while the degree of the denominator is 2.

And all that remains is to evaluate the residuals at $i$ and at $-1+ i$

At $z=i:$
$$\frac {d}{dx} \frac {z^2}{(z+i)^2(z^2+2z + 2)} = \frac{(z+i)^2(z^2+2z+2)(2z) - z^2[2(z+i)(z^2+2z+2)+(z+i)^2(2z+2)}{((z+i)^2(z^2+2z+2))^2}\Bigg|_{z=i}$$

$$\frac {(2i)^2(1+2i)(2i) - (-1)[(2)(2i)(1+2i) + (2i)^2(2i+2)}{((2i)^2(1+2i))^2}$$

$$\frac {-4i(1+2i) -4(2i+2)}{-4(1+2i)^2}$$
$$\frac {-12i}{16(-3+4i)} = \frac {-12+9i}{100}$$

And at $z = -1+i$

\begin{align*} \frac {z^2}{(z^2+1)^2(z+1+i)}\Bigg|_{z=-1+i} &=\frac {(-1+i)^2}{((-1+i)^2 +1)^2(2i)}\\ &=\frac {-2i}{(-2i+1)^2(2i)}\\ &=\frac {1}{3+4i}\\ &=\frac {3-4i}{25} \end{align*}

Now that we have our residuals...

$$\oint f(z) d\gamma = 2\pi i(\frac {-12 + 9i}{100} + \frac {3-4i}{25}) = \frac {7\pi}{50}$$