Whose integrals is it?

Start with the integral $\int_0^\infty \frac{1 - \cos ax}{x^2} \; = \; \tfrac12\pi a \hspace{1cm} a > 0$ Using the Binomial Theorem, for any positive integer $k$ , $\begin{aligned} \sin^{2k}x & = \left(\frac{e^{ix} - e^{-ix}}{2i}\right)^k \; = \; \frac{(-1)^k}{2^{2k}}\sum_{j=0}^{2k} (-1)^j {2k \choose j} e^{2(k-j)ix} \\ & = \frac{1}{2^{2k-1}}\sum_{j=1}^k (-1)^{j-1}{2k \choose k+j} (1 - \cos 2jx) \end{aligned}$ and hence $\begin{aligned} \int_0^\infty \sin^{2k}\tfrac{1}{x}\,dx & = \int_0^\infty \frac{\sin^{2k}x}{x^2}\,dx \\ & = \frac{1}{2^{2k-1}}\sum_{j=1}^k (-1)^{j-1} {2k \choose k+j} \int_0^\infty \frac{1 - \cos 2jx}{x^2}\,dx \\ & = \frac{\pi}{2^{2k-1}}\sum_{j=1}^k (-1)^{j-1} j {2k \choose k+j} \; = \; \frac{\pi}{2^{2k-1}}{2k-2 \choose k-1} \end{aligned}$ and hence $\begin{aligned} \sum_{k=1}^{N+1} \int_0^\infty \sin^{2k}\tfrac{1}{x}\,dx & = \; \pi\sum_{k=0}^N \frac{1}{2^{2k+1}}{2k \choose k} \; = \; \frac{\pi (2N+1)!!}{2^{N+1}N!} \; = \frac{\sqrt{\pi}\Gamma\big(N + \frac32\big)}{\Gamma(N+1)} \end{aligned}$ Putting $N=2017$ gives the answer $\frac{\sqrt{\pi}\Gamma\big(\tfrac{4037}{2}\big)}{\Gamma(2018)}$ and hence the answer is $1 + 4037 + 2 + 2018 = \boxed{6058}$

Alternatively, we can match the original integrals with the more standard Wallis integrals as follows: $\begin{aligned} \int_0^\infty \sin^{2k} \tfrac{1}{x}\,dx & = \int_0^\infty \frac{\sin^{2k}x}{x^2}\,dx \; = \; \tfrac12\int_{-\infty}^\infty \frac{\sin^{2k}x}{x^2}\,dx \\ & = \tfrac12\sum_{n \in \mathbb{Z}} \int_0^\pi \frac{\sin^{2k}x}{(x + n\pi)^2}\,dx \; = \; \tfrac12 \int_0^\pi \sin^{2k}x \,\mathrm{cosec}^2x\,dx \\ & = \tfrac12\int_0^\pi \sin^{2k-2}x\,dx \; = \; \int_0^{\frac12\pi}\sin^{2k-2}x\,dx \; =\; \frac{\pi}{2^{2k-1}}{2k-2 \choose k-1} \end{aligned}$