2018 times

Relevant wiki: Beta Function

$\begin{aligned} I & = \int_{-\infty}^\infty \frac 1{(x^2+1)^{2018}}\ dx & \small \color{#3D99F6} \text{Since the integrand is even} \\ & = 2\int_0^\infty \frac 1{(1+x^2)^{n+1}}\ dx & \small \color{#3D99F6} \text{Let }n = 2017 \\ & = \int_0^\infty \frac {u^{-\frac 12}}{(1+u)^{n+1}}\ dx & \small \color{#3D99F6} \text{Let }u = x^2 \implies du = 2x \ dx \\ & = B \left(\frac 12, \frac {2n+1}2\right) & \small \color{#3D99F6} \text{Beta function }B(j,k) = \int_0^\infty \frac {u^{j-1}}{(1+u)^{j+k}} \\ & = \frac {\Gamma \left(\frac 12 \right)\Gamma \left(\frac {2n+1}2 \right)}{\Gamma (n+1)} & \small \color{#3D99F6} B(j,k) = \frac {\Gamma(j)\Gamma(k)}{\Gamma(j+k)} \\ & = \frac {\sqrt \pi \left(\frac {(2n-1)!!}{2^n}\sqrt \pi \right)}{n!} & \small \color{#3D99F6} \text{where }\Gamma(\cdot) \text{ is the gamma function.} \\ & = \frac {(2n)!\pi}{\left(2^nn!\right)^2} & \small \color{#3D99F6} \text{As } (2n-1)!! = \frac {(2n)!}{2^n n!} \end{aligned}$

Therefore, $a+b+c = 2n+2+n = 3(2017) + 2 = \boxed{6053}$ .

Before to evaluate let's reduce the integral in the form of $I_n =\int \dfrac{1}{\,(a^2+x^2)^{n+1} }\,dx\qquad \forall n \in\mathbb N\,.$ 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 $n=2017$ and $a=1$ we obtain $I_{2018} = \dfrac{1}{4036} \cdot \dfrac{x}{\,(1+x^2)^{2018}}+\dfrac{4033}{4034} \cdot I_{2017}$ Now setting limits from $-\infty$ to $\infty$ We obtain the following $\begin{aligned} \int_{-\infty}^{\infty} I_{2018} & =0+\dfrac{4033}{4032}\int_{-\infty}^{\infty} I_{2017} = 0 +\dfrac{4033\cdot 4031}{4034\cdot 4032}\int_{-\infty}^{\infty} I_{2016 } \\& = 0 +\dfrac{4033\cdot 4031\cdots 1}{4036\cdot 4034\cdots 2 }\int_{-\infty}^{\infty} I_1 =\dfrac{\,(4033)!!}{(4034)!!}\pi\end{aligned}$ Since $\dfrac{\,(2n-1)!!}{(2n)!!} = \dfrac{(2n)!}{(2^n n!)^2}=\dfrac{4034!\cdot \pi}{\,(2^{2017} \cdot 2017!)^2}$ Thus, $a+b+c =4034+2017+2=\boxed{6053}$ .