# Evaluating $\int_{0}^{\infty} \left[\left(\frac{2015}{2015+x}+\cdots +\frac{2}{2+x}+\frac{1}{1+x}-x\right)^{2016}+1 \right] ^{-1}\mathrm{d}x$

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.

• Around $x =1$, $(x^{2016}-1)= (x-1)(1+\cdots)$, so isn't there an issue with the easier integral? So should the - be a +? Dec 22, 2015 at 13:25
• Is this an early Xmas present from your friend ? Dec 22, 2015 at 13:33
• The two integrals are not equal, the first one is around $1.366$, the second one is around $1$. maybe you want to replace the bound of the first integral from $\int_0^\infty$ to $\frac12\int_{-\infty}^\infty$ ... Dec 22, 2015 at 13:46
• The reason I suggest replacing the limit of first integral from $\int_0^\infty$ to $\frac12\int_{-\infty}^\infty$ because the statement will become true. Alternatively, the two integral will be equal if you replace both lower limit from $0$ to $-\infty$. Dec 22, 2015 at 14:05
• These problems usually come from general statements, so I would expect this one to be valid for a large class of natural numbers, not just for $2015$. Nevertheless, (superficially) trying it for $1$ doesn't seem to work. Dec 22, 2015 at 14:06

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\int_{0}^{\infty}\bracks{\pars{{2015 \over 2015 + x} + \cdots + {2 \over 2 + x} + {1 \over 1 + x} - x}^{2016} + 1}^{-1}\,\dd x:\ {\large ?}}$.

\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.

• +1 cool, I never thought of actually approximate the integral analytically. Jun 19, 2017 at 17:39
• @achillehui Thanks. Jun 19, 2017 at 19:20
• +1 Thanks for a much more accurate and satisfying computation. :) Gave you the bounty. Jun 24, 2017 at 3:04
• @Kugelblitz Thanks a lot. $\left(\,\substack{\bullet\quad\bullet\\ \mid\\[2mm] \smile\\[0.5mm]\smile}\,\right)$ Jun 24, 2017 at 3:12

Edited: This integral evaluation is of a related integral which is equivalent to $\int^{\infty}_0(x^{2016}+1)^{-1}dx$, for the original integral in the question, differs in evaluation for limits from $0$ to $\infty$ owing to symmetry considerations discussed in the comments.

$$I=\frac{1}{2}\int_{-\infty}^{\infty} \left[\left(\frac{2015}{2015+x}+\cdots +\frac{2}{2+x}+\frac{1}{1+x}-x\right)^{2016}+1 \right] ^{-1}\mathrm{d}x$$ $$I=\frac{1}{2}\int_{-\infty}^{\infty} \left[\left(\sum^{2015}_{i=1}\frac{i}{x+i}-x\right)^{2016}+1 \right] ^{-1}\mathrm{d}x$$

Now, letting $f(x)=\frac{1}{x^{2016}+1}$, and noting that $f(x)=f(-x)$, $$I=\frac{1}{2}\int_{-\infty}^{\infty} f\left(\sum^{2015}_{i=1}\frac{i}{x+i}-x\right)\mathrm{d}x=\frac{1}{2}\int_{-\infty}^{\infty} f\left(-\left(\sum^{2015}_{i=1}\frac{i}{x+i}-x\right)\right)\mathrm{d}x$$ $$I=\frac{1}{2}\int_{-\infty}^{\infty} f\left(x-\sum^{2015}_{i=1}\frac{i}{x-(-i)}\right)\mathrm{d}x \tag {1}$$ Using Glasser's Master Theorem, $$I=\frac{1}{2}\int^{\infty}_{-\infty} f(x)\ \mathrm{d}x=\frac{1}{2}\int^{\infty}_{-\infty} \frac{1}{x^{2016}+1}\ \mathrm{d}x=\int^{\infty}_{0} \frac{1}{x^{2016}+1}\ \mathrm{d}x \tag {2}$$

Now we know that $$B(a,b)=\int^{\infty}_0\frac{t^{a-1}}{(1+t)^{a+b}}dt$$ From $(2)$,after substituting $x^{2016} =t$, $$I=\frac{1}{2016}\int^{\infty}_{0}\frac{t^{\frac{1}{2016}-1}}{(1+t)^{\frac{1}{2016}+\frac{2015}{2016}}}dt=\frac{1}{2016}B(\frac{1}{2016},\frac{2015}{2016})$$ Therefore $$\color{red}{I=\frac{1}{2016}\frac{\Gamma(\frac{1}{2016})\Gamma(\frac{2015}{2016})}{\Gamma(1)}=\frac{\pi}{2016\sin(\frac{\pi}{2016})}\approx1.0000004047320180811575}$$

• Next you can put $x^{2016} =t$ and evaluate in terms of Beta function namely $B(1/2016,2015/2016)/2016=\pi/(2016\sin(\pi/2016))$ May 6, 2017 at 5:28
• In case you are wondering this downvote is not mine. I do not downvote anonymously. May 6, 2017 at 5:32
• I know. And thank you! I'll modify the answer. (Rather, add on) May 6, 2017 at 5:33
• Is $I$ symmetric to the origin? I think of course not and you should not expand the integral to negative reals... May 6, 2017 at 7:19
• It is not symmetrical. If you compute the sum $p(x) =\left( \sum_{k=1}^{2015}\frac{k}{k+x} \right)- x$ at $x = \pm \frac12$, you will notice $p(\frac12) \approx 1003.7, p(-\frac12) \approx 3027.3$ This implies $\frac{1}{p(x)^{2016} + 1}$ takes different value at $x = \pm \frac12$. Jun 17, 2017 at 7:21