# 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 $\frac{\arctan (x)}{x}$?

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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.