TITLE: Evaluating $\int_{0}^{\infty} \left[\left(\frac{2015}{2015+x}+\cdots +\frac{2}{2+x}+\frac{1}{1+x}-x\right)^{2016}+1 \right] ^{-1}\mathrm{d}x$
QUESTION [13 upvotes]: I need to evaluate $$\int_{0}^{\infty} \left[\left(\frac{2015}{2015+x}+\cdots  +\frac{2}{2+x}+\frac{1}{1+x}-x\right)^{2016}+1 \right] ^{-1}\mathrm{d}x
$$
I've been told that the way forward is showing that the integral is the same as $$\int_0^{\infty} (x^{2016} + 1)^{-1} \, \mathrm{d}x$$
i.e: that the weird sum of fractions doesn't affect the integral. 
I've tried $$\sum_{n=1}^{2015} \frac{n}{n+x} = \sum_{n=1}^{2015} \left(1 - \frac{x}{n+x}\right) = 2015 - \sum_{n=1}^{2015} \frac{x}{n+x}$$
but it's getting me nowhere.

REPLY [12 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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$\ds{\int_{0}^{\infty}\bracks{\pars{{2015 \over 2015 + x} + \cdots +
{2 \over 2 + x} + {1 \over 1 + x} - x}^{2016} + 1}^{-1}\,\dd x:\ {\large ?}}$.

\begin{align}
&\int_{0}^{\infty}\bracks{\pars{{2015 \over 2015 + x} + \cdots +
{2 \over 2 + x} + {1 \over 1 + x} - x}^{2016} + 1}^{-1}\,\dd x
\\[5mm] = &\
\int_{0}^{\infty}{\dd x \over \mc{F}^{2016}\pars{x} + 1}\quad
\mbox{where}\quad\mc{F}\pars{x} \equiv \sum_{k = 1}^{2015}{k \over k + x} - x
\end{align}

\begin{align}
\mc{F}\pars{x} & \equiv 
\sum_{k = 1}^{2015}{k \over k + x} - x =
2015 - x\sum_{k = 1}^{2015}{1 \over k + x} - x
\\[5mm] & =
2015 - x\sum_{k = 1}^{\infty}\pars{{1 \over k + x} - {1 \over k + 2015 + x}} - x \\[5mm] & =
2015 - x\pars{H_{x + 2015} - H_{x}} - x\qquad
\pars{~H_{z}:\ Harmonic\ Number~}
\end{align}

\begin{align}
&\bbx{\mc{F}\pars{x} =
2015 - \pars{\vphantom{\large A}H_{x + 2015} - H_{x} + 1}x}
\\[5mm]
&\mbox{Some characteristic behaviours of}\ \mc{F}\pars{x}\ \mbox{are}\
\left\{\begin{array}{l}
\ds{\mc{F}\pars{0} = 2015}
\\[2mm]
\ds{\mc{F}\pars{x} \to -\infty\quad \mbox{as}\quad x \to \infty}
\\[2mm]
\ds{\mc{F}'\pars{x} \leq 0\,,\quad \forall\ x \geq 0}
\\[2mm]
\ds{\mc{F}\pars{r} = 0\,,\quad r \approx 939.105}
\\[2mm]
\ds{\mc{F}\pars{939.105} \approx -6.28337 \times 10^{-4}}
\\[2mm]
\ds{\mc{F}'\pars{939.105} \approx -1.46404}
\end{array}\right.
\\[1cm]
&\mbox{Hereafter, I'll perform a numerical evaluation which is based in the Laplace Method}:
\\
&\int_{0}^{\infty}{\dd x \over \mc{F}^{2016}\pars{x} + 1} =
\int_{0}^{r}{\dd x \over \mc{F}^{2016}\pars{r - x} + 1} +
\int_{0}^{\infty}{\dd x \over \mc{F}^{2016}\pars{x + r} + 1}
\\[5mm] \approx &\
2\int_{0}^{\infty}
\expo{-\bracks{\mc{F}'\pars{r}}^{\large 2016}x^{\large 2016}}\,\dd x =
{2 \over \verts{\mc{F}'\pars{r}}}\int_{0}^{\infty}\expo{-x^{\large 2016}}\,\dd x
\\[5mm] = &\
{2 \over \verts{\mc{F}'\pars{r}}}\,{1 \over 2016}
\int_{0}^{\infty}x^{1/2016 - 1}\expo{-x}\,\dd x =
{1 \over 1008}\,{1 \over \verts{\mc{F}'\pars{r}}}\,\Gamma\pars{1 \over 2016}
\\[5mm] & \approx \bbx{1.36569}
\quad\mbox{with}\quad
\left.\mc{F}'\pars{r}\right\vert_{\ r\ \approx\ 939.105} \approx -1.46404
\end{align}

This value $\ds{\pars{~1.36569~}}$ is the numerical one reported by $\texttt{@achille hui}$ in another answer comment.