Trouble with integrating $\frac{\arctan (x)}{x}$ 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 $\frac{\arctan (x)}{x}$?
 A: Try to find a way to solve the problem without doing the integration. What do you know about the arctangent as $x\to\infty$?
A: 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.
A: $\newcommand{\+}{^{\dagger}}%
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\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 $\pars{1}$.
