Just another Integral

Let

$I=\displaystyle\int_0^{2017}\dfrac{e^{\cos(\pi\{x\})}}{e^{\cos(\pi\{x\})}+e^{-\cos(\pi\{x\})}}dx$

Since the function $\{x\}$ is periodic with period $1$ , we can rewrite the integral as follows

$I=2017\displaystyle\int_0^1\dfrac{e^{\cos(\pi x)}}{e^{\cos(\pi x)}+e^{-\cos(\pi x)}}dx$

.

We now have to evaluate

$I_1=\displaystyle\int_0^1\dfrac{e^{\cos(\pi x)}}{e^{\cos(\pi x)}+e^{-\cos(\pi x)}}dx$

Via the substitution $u=\pi x,dx=\dfrac{du}{\pi}$

$I_1=\dfrac{1}{\pi}\displaystyle\int_0^{\pi}\dfrac{e^{\cos(u)}}{e^{\cos(u)}+e^{-\cos(u)}}du$

Using the integration trick

$\displaystyle\int_a^b f(x)=\displaystyle\int_a^b f(a+b-x)$

$\begin{aligned} \Longrightarrow I_1&=\dfrac{1}{\pi}\displaystyle\int_0^{\pi}\dfrac{e^{\cos(u)}}{e^{\cos(u)}+e^{-\cos(u)}}du\\ &=\dfrac{1}{\pi}\displaystyle\int_0^{\pi}\dfrac{e^{\cos(\pi-u)}}{e^{\cos(\pi-u)}+e^{-\cos(\pi-u)}}du\\ &=\dfrac{1}{\pi}\displaystyle\int_0^{\pi}\dfrac{e^{-\cos(u)}}{e^{-\cos(u)}+e^{\cos(u)}}du \end{aligned}$

Adding up

$\begin{aligned} 2I_1&=\dfrac{1}{\pi}\displaystyle\int_0^{\pi}\dfrac{e^{\cos(u)}+e^{-\cos(u)}}{e^{\cos(u)}+e^{-\cos(u)}}du\\ &=\dfrac{1}{\pi}\displaystyle\int_0^{\pi} du\\ &=\dfrac{1}{\pi}\cdot \pi=1\Longrightarrow I_1=\dfrac{1}{2} \end{aligned}$

Hence

$I=2017I_1=\dfrac{2017}{2}=\dfrac{a}{b}$

$\Longrightarrow a+b=2017+2=\boxed{2019}$