Complex analysis on real integral Use complex analysis to compute the real integral
$$\int_{-\infty}^\infty \frac{dx}{(1+x^2)^3}$$
Attempt
I think I want to consider this as the real part of
$$\int_{-\infty}^\infty \frac{dz}{(1+z^2)^3}$$
and then apply the residue theorem. However, I am not sure how that is the complex form and the upper integral is the real part and how to apply.
 A: At a first glance this looks like something that can be solved using the method of contour integration. If we consider a semicircle in the upper half plane with radius $R$, we see that in this region, the function $\dfrac{1}{(1+z^2)^3}$ has a pole at $z=i$, so the residue theorem states that the integral taken "around" the semicircle should be $2\pi i$ times the residue at $z=i$.
If you can then compute the integral around the semicircle "without" the real axis, you should be able to get the value of your integral, as we let $R\to \infty$.
Does this help?
A: In order to perform such an integration, you need to specify a specific contour over which you're integrating, since the singularities of the integrating function vary.
A standard way of solving such integrals is :
The singularities of the integrating function : 
$$f(x) = \frac{1}{(1+x^2)^3}$$
come as : $x^2 + 1 = 0 \Rightarrow x=\pm i$.
We will integrate the function $f(x)$ with respect to $x$, over the partially smooth curve that is consisted of the half-moon of the upper level of $γ_R$, with $z(θ) = Re^{iθ}$, $0 \geq θ \geq π$ and the line segment $[-R,R]$, where we take $R$ to be big enough, so that the singularity points $(x=i)$ in the upper plane are inside $γ_R$. 
From the Residue-Theorem we have :
$$\int_{-R}^R \frac{1}{(1+x^2)^3}\,dx + \int_{γ_R}\frac{1}{(1+x^2)^3} \, dz = 2\pi i\operatorname{Res}(f(x),i)$$ 
The function also satisfies Jordan's Lemma, so it is : 
$$\lim_{R\to \infty}\int_{γ_R}\frac{1}{(1+x^2)^3}=0$$
thus :
$$ \int_{-\infty}^\infty\frac{1}{(1+x^2)^3} \, dx=\lim_{R\to \infty} \int_{-R}^R \frac{1}{(1+x^2)^3} \, dx= 2\pi i \operatorname{Res}(f(x),i)$$
I'll leave the residue calculation to you!
A: It is much faster to apply Feynman's trick. For any $a>0$ we clearly have
$$ \int_{-\infty}^{+\infty}\frac{dx}{a+x^2} = \frac{\pi}{\sqrt{a}} \tag{1} $$
and by applying $\frac{d^2}{da^2}$ to both sides:
$$ \int_{-\infty}^{+\infty}\frac{2\,dx}{(a+x^2)^3} = \frac{3\pi}{4a^2\sqrt{a}} \tag{2} $$
hence by evaluating $(2)$ at $a=1$ we instantly get
$$ \int_{-\infty}^{+\infty}\frac{dx}{(1+x^2)^3}\,dx = \frac{3\pi}{8}\tag{3}$$
and in general
$$ \forall m\geq 1,\qquad \int_{-\infty}^{+\infty}\frac{dx}{(1+x^2)^{m}}\,dx = \frac{\pi}{4^{m-1}}\binom{2m-2}{m-1}.\tag{4}$$
A: Define a contour that is a semicircle in the upper half plane of radius R.  Plus the real line from $-R$ to $R$
Then let R get to be arbitrarily large.
There is one pole at $z = i$ inside the contour
Cauchy integral formula says:
$$f^{(n)}(a) = \frac {n!}{2\pi i}\oint \frac {f(a)}{(z-a)^{n+1}}\ dz$$
$$\oint \frac {\frac {1}{(z+i)^3}}{(z-i)^3}\ dz = \pi i \frac{d^2}{dz^2} \frac {1}{(z+i)^3} \text{ evaluated at } z=i.$$
Next you will need to show that the integral along contour of the semi-cricle goes to $0$ as $R$ gets to be large.  
$$z = R e^{it}, dz = iR e^{it}\\
\displaystyle\int_0^{\pi} \frac {iRe^{it}}{(R^2 e^{2it} + 1)^3} \ dt\\
\left|\frac {iRe^{it}}{(R^2 e^{2it} + 1)^3}\right| < R^{-5}\\
\left|\int_0^{\pi} \frac {iRe^{it}}{(R^2 e^{2it} + 1)^3} \ dt\right| < \int_0^{\pi} R^{-5}\ dt\\
\lim_\limits{R\to \infty}\int_0^{\pi} R^{-5}\ dt = 0$$
A: First we try to make a complex integral. Suppose we wanna solve:
$$\int{dz\over{(1+z^2)^3}}$$
when $z$ moves on curvature $C$ with below definition:
$C:$
$-R\le z\le R$ when $z\in \Bbb R$
$$z=Re^{i\theta} \text{ and } 0\le \theta\le\pi$$
i.e. $C$ is a semicircle of radius $R$ on the positive half-plane of $\Bbb C$. then we can break the complex integral to two as following:
$$ \int{dz\over{(1+z^2)^3}}=\int_{-R}^{R}{dx\over{(1+x^2)^3}}+\int_{0}^{\pi}{{iRe^{i\theta}d\theta}\over{(1+R^2e^{2i\theta})^3}}$$
Using complex analysis:
$$\int_{C}{dz\over{(1+z^2)^3}}={2\pi i\Sigma_{i=1}^{n}{Res_{C}(z_i)}}$$
where $Res(z_i)$ indicates on residue of $f(z)={1\over{(1+z^2)^3}}$ in singularity $z_i$ included by $C$. Since for our function the only such singularity is $z=i$ for sufficiently large $R$ we the obtain:
$$\int_{C}{dz\over{(1+z^2)^3}}={2\pi i\Sigma_{i=1}^{n}{Res_{C}(i)}}$$
where $Res_{C}(i)$ has been calculated as follows:
$$Res_{C}(i)={1\over 2!}{lim_{z\to i}{{{d^2}\over{dz^2}}}{{(z-i)^3}{1\over{(1+z^2)^3}}}}={{-3\over 16}i}$$
since the singularity is of order 3 by factorizing ${1\over{(1+z^2)^3}}$ to ${1\over{(z+i)^3(z-i)^3}}$. Then the complex integral would be:
$$\int_{C}{dz\over{(1+z^2)^3}}=(2\pi i)({{-3\over 16}i})={3\pi\over 8}$$
With $R\to\infty$ the second term of right side tends to zero, because order of $R$ in denominator is greater than that of numerator therefore:
$$\int_{-\infty}^\infty {dx\over{(1+x^2)^3}}={3\pi \over 8}$$
