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I have a function $F(x)$ that is defined as $\int_0^x f(t) dt$ which I'm trying to find the limit of when $x$ approaches infinity. Previously in the assignment, the function $f(x)$ was defined as being $$ f(x) = \begin{cases} \frac{\arctan (x)}{x}, &x \neq 0, \\ 1, &x = 0. \end{cases} $$ I'm having trouble integrating $\frac{\arctan (x)}{x}$. when I try to see the result in Wolfram Alpha, I get a result with imaginary numbers and polylogarithms both of which aren't part of our course.

So I was wondering if I have misunderstood the task completely, or if there is some trick to integrating $\frac{\arctan (x)}{x}$?

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up vote 6 down vote accepted

Try to find a way to solve the problem without doing the integration. What do you know about the arctangent as $x\to\infty$?

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the limit of the arctangent as x approaches infinity is pi divided by two. So the limit of $(arctan x)/x$ is 0 since x is much greater than arctan x. So the integral of that has to be infinite if i'm not mistaken, but i'm not sure how sound that reasoning is. – Kracobsen Nov 1 '11 at 22:16
@Kracobsen: why is the integral of that infinite? In fact, the fact that the limit as $x$ goes to infinity of $f(x)$ (in this case, using $f(x) = \arctan(x)/x$) is 0 is a necessary condition for the integral to not be infinite! But it's not a sufficient condition - that is, even if $f(x)$ goes to 0, its integral can still be infinite; you need to justify that that's what happens here. – Steven Stadnicki Nov 1 '11 at 22:25
If $\lim_{x \to \infty} f(x) = 0$, then $\lim_{x \to \infty} \int_0^x f(t) \: dt$ can be finite or infinite. Consider the cases $f(t) = 1/(1+t^2)$ (finite integral) and $f(t) = 1/(1+t)$ (infinite integral). Which of these cases does $f(t) = \arctan(t)/t$ resemble more closely? – Michael Lugo Nov 1 '11 at 22:28
Do you know about the comparison test for convergence of infinite series? Do you reckon there might be some similar technique for improper integrals? – Gerry Myerson Nov 2 '11 at 3:32

Use Taylor series representation for $\arctan$: $$ \frac{\arctan(x)}{x} = \sum_{k=0}^\infty \frac{(-1)^{k}}{2k+1} x^{2k} $$ and integrate term-wise, for $0<x<1$: $$ \begin{eqnarray} \int_0^x \frac{\arctan(t)}{t} \mathrm{d} t &=& \sum_{k=0}^\infty \frac{(-1)^{k}}{(2k+1)^2} x^{2k+1} = x \sum_{k=0}^\infty \frac{(-x^2)^{k}}{k!} \frac{\left(\frac{1}{2}\right)_k \cdot \left(\frac{1}{2}\right)_k \cdot k!}{\left(\frac{3}{2}\right)_k \cdot \left(\frac{3}{2}\right)_k} \\ &=& x \cdot {}_3 F_2\left(\frac{1}{2},\frac{1}{2}, 1; \frac{3}{2}, \frac{3}{2}; -x^2 \right) \end{eqnarray} $$ wher $(a)_k := a(a+1)\cdots(a+k-1)$ denotes Pochhammer symbol and ${}_pF_q$ denotes generalized hypergeometric function. For $x>1$, we split the integration range $(0,x)$ into $(0,1)$ and $(1,x)$ and perform a change of variables $t \to 1/t$ in the second integral: $$\begin{eqnarray} \int_0^x \frac{\arctan(t)}{t} \mathrm{d} t &=& \int_0^1 \frac{\arctan(t)}{t} \mathrm{d} t + \int_1^{x} \frac{\arctan(t)}{t} \mathrm{d} t \\ &=& \int_0^1 \frac{\arctan(t)}{t} \mathrm{d} t + \int_{1/x}^1 \frac{\arctan(1/t)}{t} \mathrm{d} t \end{eqnarray} $$ Using the identity $\arctan(t) + \arctan(1/t) = \frac{\pi}{2}$ which holds for $t>0$, we have: $$\begin{eqnarray} \int_0^x \frac{\arctan(t)}{t} \mathrm{d} t &=& \int_0^1 \frac{\arctan(t)}{t} \mathrm{d} t + \int_1^{x} \frac{\arctan(t)}{t} \mathrm{d} t \\ &=& \int_0^{1/x} \frac{\arctan(t)}{t} \mathrm{d} t + \frac{\pi}{2} \int_{1/x}^1 \frac{1}{t} \mathrm{d} t \\ &=& \frac{\pi}{2} \ln(x) + \frac{1}{x} \cdot {}_3 F_2\left(\frac{1}{2},\frac{1}{2}, 1; \frac{3}{2}, \frac{3}{2}; -\frac{1}{x^2} \right) \end{eqnarray} $$ As a corollary, $\int_0^x \frac{\arctan(t)}{t} \mathrm{d}t$ diverges logarithmically as $x$ grows large. Of course, this can be determined by much simpler means, as in Gerry's answer.

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \int_{0}^{x}{\arctan\pars{t} \over t}\,\dd t&=\ln\pars{x}\arctan\pars{x} -\int_{0}^{x}{\ln\pars{t} \over t^{2} + 1}\,\dd t \\[3mm]&=\ln\pars{x}\arctan\pars{x} -\lim_{\mu\to 0}\partiald{}{\mu}\int_{0}^{x}{t^{\mu} \over t^{2} + 1}\,\dd t \\[3mm]&=\ln\pars{x}\arctan\pars{x} -\lim_{\mu\to 0}\partiald{}{\mu}\int_{0}^{x}{t^{\mu/2} \over t + 1}\,\half\,t^{-1/2}\,\dd t \\[3mm]&=\ln\pars{x}\arctan\pars{x} - \half\,\lim_{\mu\to 0}\partiald{}{\mu} \int_{0}^{x}{t^{\pars{\mu - 1}/2} \over t + 1}\,\dd t\tag{1} \end{align}

With $\ds{\xi \equiv {1 \over t + 1}\quad\iff\quad t = {1 \over \xi} - 1} = {1 - \xi \over \xi}$ \begin{align} \int_{0}^{x}{t^{\pars{\mu - 1}/2} \over t + 1}\,\dd t&= \int_{1}^{1/\pars{x + 1}}\xi\pars{1 - \xi \over \xi}^{\pars{\mu - 1}/2}\, \pars{-\,{\dd \xi \over \xi^{2}}} \\[3mm]&= \int^{1}_{1/\pars{x + 1}}\xi^{-\pars{\mu + 1}/2}\pars{1 - \xi}^{\pars{\mu - 1}/2}\, \dd \xi\\[3mm]&={\rm B}\pars{{1 - \mu \over 2},{1 + \mu \over 2}} -{\rm B}_{1/\pars{x + 1}}\pars{{1 - \mu \over 2},{1 + \mu \over 2}}\tag{2} \end{align} where ${\rm B}\pars{p,q}$ and ${\rm B}_{x}\pars{p,q}$ are the Beta and the Incomplete Beta functions, respectively. ${\rm B}\pars{p,q}$ satisfies $\ds{{\rm B}\pars{p,q} = {\Gamma\pars{p}\Gamma\pars{q} \over \Gamma\pars{p +} q}}$ such that $$ {\rm B}\pars{{1 - \mu \over 2},{1 + \mu \over 2}} ={\Gamma\pars{1/2 - \mu/2}\Gamma\pars{1/2 + \mu/2} \over \Gamma\pars{1}} ={\pi \over \sin\pars{\pi\bracks{1/2 + \mu/2}}} ={\pi \over \cos\pars{\pi\mu/2}} $$ $\pars{2}$ is reduced to: $$ \int_{0}^{x}{t^{\pars{\mu - 1}/2} \over t + 1}\,\dd t =\pi\sec\pars{\pi\mu \over 2}-{\rm B}_{1/\pars{x + 1}}\pars{{1 - \mu \over 2},{1 + \mu \over 2}} $$ In terms of the Hipegeometric Function: $$ \int_{0}^{x}{t^{\pars{\mu - 1}/2} \over t + 1}\,\dd t =\pi\sec\pars{\pi\mu \over 2} - 2\,{\pars{1 + x}^{\pars{\mu - 1}/2} \over 1 - \mu}\ _{2}{\rm F}_{1}\pars{{1 - \mu \over 2},{1 - \mu \over 2};{3 - \mu \over 2}; {1 \over x + 1}} $$ and \begin{align} &\lim_{\mu \to 0}\partiald{}{\mu} \int_{0}^{x}{t^{\pars{\mu - 1}/2} \over t + 1}\,\dd t \\[3mm]&=-\,\bracks{\ln\pars{1 + x} + 2}\arcsin\pars{1 \over \root{1 + x}} \\[3mm]&\phantom{=}- \\[3mm]&\phantom{=} {2 \over \root{1 + x}}\lim_{\mu \to 0}\partiald{}{\mu}\ _{2}{\rm F}_{1}\pars{{1 - \mu \over 2},{1 - \mu \over 2};{3 - \mu \over 2}; {1 \over x + 1}} \end{align}

The answer is found by replacing this expression in $\pars{1}$.

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