Just Another Ole Integral

Relevant wiki: Beta Function

$\begin{aligned} I & = \int_{-\infty}^\infty \frac 1{(x^2+1)^4} dx & \small \color{#3D99F6} \text{Since the integrand is even} \\ & = 2 \int_0^\infty \frac 1{(x^2+1)^4} dx & \small \color{#3D99F6} \text{Let }u = x^2 \implies du = 2x\ dx \\ & = \int_0^\infty \frac {u^{-\frac 12}}{(1+u)^4} du & \small \color{#3D99F6} \text{Beta function }B(m,n) = \int_0^\infty \frac {u^{m-1}}{(1+u)^{m+n}} du \\ & = B \left(\frac 12, \frac 72\right) & \small \color{#3D99F6} \text{and }B(m,n) = \frac {\Gamma (m)\Gamma (n)}{\Gamma(m+n)}, \\ & = \frac {\Gamma\left(\frac 12\right)\Gamma\left(\frac 72\right)}{\Gamma(4)} & \small \color{#3D99F6} \text{where }\Gamma (\cdot) \text{ is gamma function.} \\ & = \frac {\sqrt \pi \left(\frac 52 \times \frac 32 \times \frac 12 \sqrt \pi \right)}{3!} \\ & = \frac {5\pi}{16} \end{aligned}$

Therefore, $a+b+c = 1+1+4 = \boxed 6$ .

Before to evaluate let's reduce the integral in the form of $I_n \int \dfrac{1}{\,(a^2+x^2)^{n+1}}\,dx\qquad \forall n \in\mathbb N \geq 0 \,.$ Here we will integrate it by parts so let us make $u$ substitution as $u = \dfrac{1}{\,(a^2+x^2)^n}$ and $\,dv =\,du$ which follows as $\,du = -\dfrac{2n x\,dx}{\,(a^2+x^2)^{n+1}}$ and $v=x$ . Hence, $\begin{aligned} I_n & = \dfrac{x}{\,(a^2+x^2)^n } +2n \int\dfrac{x^2}{\,(a^2+x^2)^{n+1}}\,dx \\& = \dfrac{x}{\,(a^2+x^2)}+ 2n\int \dfrac{\,a^2+x^2-a^2}{\,(a^2+x^2)^{n+1} } \\& = \dfrac{x}{\,(a^2+x^2)}+2nI_n -2na^2I_{n+1} \\ \therefore I_{n+1}& =\dfrac{1}{2na^2} \cdot \dfrac{x}{\,(a^2+x^2)^n}+\dfrac{2n-1}{2na^2} \cdot I_n\end{aligned}$ Set $n=3$ and $a=1$ we obtain $I_4 = \dfrac{1}{6} \cdot \dfrac{x}{\,(1+x^2)^4}+\dfrac{5}{6} \cdot I_3$ It is known that $\begin{aligned}\int\dfrac{1}{a^2+x^2} & = \dfrac{1}{a} \tan ^{-1} \dfrac{x}{a} +C\Rightarrow\int\dfrac{1}{1+x^2} = \tan ^{-1} x +C\\ I_2 =I_{1+1} & =\dfrac{x}{2\,(1+x^2)} +\dfrac{1}{2}\tan^{-1} x +C \end{aligned}$ Therefore, $I_4 = \dfrac{1}{6} \cdot \dfrac{x}{\,(1+x^2)^4}+\dfrac{5}{6} \cdot \left(\dfrac{1}{4x\,(1+x^2)^2}+\dfrac{3x}{8\,(1+x^2)}+\dfrac{3}{8}\tan ^{-1} x+ C \right)$ Set the upper and lower limit we obtain $I_4 =\int_{-\infty}^{\infty} \dfrac{1}{\,(1+x^2)^4}\,dx =\dfrac{5}{6}\cdot \dfrac{3}{8} \cdot \dfrac{2\pi}{2} =\dfrac{5^1\cdot \pi^1 }{2^4}$ Finally, $a+b+c = \boxed{6}$ .

$\begin{aligned}I&=2\int_0^\infty\frac1{(x^2+1)^4}dx\\ &=2\int_\infty^0\frac1{(\frac1{u^2}+1)^4}\frac{-du}{u^2}&&\text{letting }x=\frac1u\\ &=2\int_0^\infty\frac1{u^2\frac{(1+u^2)^4}{u^8}}du\\ &=2\int_0^\infty\frac{u^6}{(u^2+1)^4}du\\ I&=2\int_0^\infty\frac{x^6}{(x^2+1)^4}dx\\ \therefore 2I&=2\int_0^\infty\frac{x^6+1}{(x^2+1)^4}dx&&\text{by adding }I=2\int_0^\infty\frac1{(x^2+1)^4}dx\\ I&=\int_0^\infty\frac{(x^2+1)(x^4-x^2+1)}{(x^2+1)^4}dx\\ &=\int_0^\infty\frac{(x^2+1)^2-3x^2}{(x^2+1)^3}dx\\ &=\int_0^\infty\frac{1}{x^2+1}-\frac{3x^2}{(x^2+1)^3}dx\\ &=\int_0^{\frac\pi2}\left(\frac{1}{\tan^2\theta+1}-\frac{3\tan^2\theta}{(\tan^2\theta+1)^3}\right)\sec^2\theta d\theta&&\text{letting }x=\tan\theta\\ &=\int_0^{\frac\pi2}\left(\frac{1}{\sec^2\theta}-\frac{3\tan^2\theta}{(\sec^2\theta)^3}\right)\sec^2\theta d\theta\\ &=\int_0^{\frac\pi2}1-3\sin^2\theta\cos^2\theta d\theta\\ &=\int_0^{\frac\pi2}1-\frac34(2\sin\theta\cos\theta)^2d\theta\\ &=\int_0^{\frac\pi2}1-\frac34\sin^22\theta d\theta&&\text{using }\sin2\theta=2\sin\theta\cos\theta\\ &=\int_0^{\frac\pi2}1-\frac38(1-\cos2\theta)d\theta&&\text{using }\sin^2\theta=\frac12(1-\cos2\theta)\\ &=\int_0^{\frac\pi2}\frac58+\frac38\cos2\theta d\theta\\ &=\left[\frac58+\frac3{16}\sin2\theta\right]_0^{\frac\pi2}\\ &=\frac{5\pi}{16}\end{aligned}$

Let's consider the complex valued function $\frac{1}{(z^2+1)^4}$ . This has poles at $z=\pm i$ . Let $\Gamma$ be the contour containing $z=i$ and the entire real axis. Therefore we have $\oint_{\Gamma} \frac{1}{(z^2+1)^4} \,dz =\oint_{\Gamma} \frac{\frac{1}{(z+i)^4}}{(z-i)^4}\,dz.$ The Cauchy integral formula for derivatives states $f^{(\lambda)}(z) = \frac{\lambda !}{2\pi i} \oint_{\Gamma} \frac{f(t)}{(t-z)^{\lambda +1}} \,dt$ Some simple rearranging gives $\frac{2\pi i f^{\lambda}(z)}{\lambda !} = \oint_{\Gamma} \frac{f(t)}{(t-z)^{\lambda +1}} \,dt.$ Note that in our case, $z=i$ , $\lambda = 3$ and $f^{(3)}(i) = \frac{-120}{2^7i^7} = \frac{15}{16i}$ . Therefore $\int_{-\infty}^{\infty}\frac{1}{(x^2+1)^4}\,dx = \oint_{\Gamma} \frac{\frac{1}{(z+i)^4}}{(z-i)^4}\,dz = \frac{2\pi i f^{\lambda}(z)}{\lambda !} = \frac{5\pi}{16}.$ Therefore $a+b+c = \boxed{6}$ .