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Problem: Evaluate:

$$\displaystyle\int_{0}^{\infty} \dfrac{1}{x} \left(\tan^{-1}(\pi x) - \tan^{-1}x\right)dx.$$

Attempt: $$\displaystyle\int_{0}^{\infty} \dfrac{1}{x} \left(\tan^{-1}(\pi x) - \tan^{-1}x\right)dx.$$ $$=\displaystyle\int_{0}^{\infty} \dfrac{tan^-1 (\pi x)}{x} dx -\int_0^\infty \dfrac{\tan^{-1}x}{x}dx.$$ Consider $$J(b) = \int_0^\infty \dfrac{\tan^-1(bx)}{x} dx$$ Differentiating $J(b)$ w.r.t (b) $$J'(b)=\int_0^\infty \dfrac{dx}{1+(bx)^2}=\dfrac{\pi}{2b}$$ $$\Longrightarrow J(b) = \dfrac{\pi}{2}\ln b + C$$ Now how do we proceed further to find C? $J(0) = 0$, but $\ln(0)$ is not defined. $$$$

User1234
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    For $b > 0$ your $J(b)$ is meaningless: the integral does not converge. There's no problem near $x = 0$, where $\arctan(bx)/x \approx b - b^3x^2/3$, but out near $x = \infty$ the value of the integrand behaves like $(\pi/2)/x$, so the integral diverges since $1/x$ is not integrable out towards $\infty$. The original question is about a difference under the integral sign, where convergence is not problematic. What you've done is similar to studying a convergent series by writing it as a difference of two divergent series: a dangerous move unless you are more careful than you have been. – KCd Jun 01 '15 at 03:12
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    You split your convergent integral into two divergent ones. Don't do that. Let $J(b)$ be the original integral with $\pi$ replaced by $b$. Then you will find that $J(b)=(\pi/2)\ln b+c$, and $b=1$ trivially gives $0=J(1)=c$. Thus, $J(\pi)=(\pi/2)\ln\pi$. – mickep Jun 01 '15 at 04:32
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    @BetterWorld, review the meaning of improper integrals. If you did not understand my explanation for why $\int_0^\infty \arctan(bx)/x ,dx$ diverges then please speak to a calculus teacher in person. Your treatment is like studying the convergent series $\sum_{n \geq 1} (-1)^{n-1}/n = 1 - 1/2 + 1/3 - 1/4 + \cdots$ by writing it as $S_1 - S_2$, where $S_1 = 1 + 1/3 + \cdots = \sum_{m \geq 1} 1/(2m+1)$ and $S_2 = 1/2 + 1/4 + \cdots = \sum_{m \geq 1} 1/(2m)$. Both $S_1$ and $S_2$ diverge. – KCd Jun 01 '15 at 10:53
  • The $J(b)$ you define diverges, because the integrand (is positive and) goes to zero too slowly. – mickep Jun 01 '15 at 10:59
  • Correct! (Both integrals are divergent.) – mickep Jun 01 '15 at 11:55

3 Answers3

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$$\begin{eqnarray*}\int_{0}^{+\infty}\frac{\arctan(\pi x)-\arctan x}{x}\,dx &=& \int_{0}^{+\infty}\frac{1}{x}\int_{1}^{\pi}\frac{x}{1+a^2 x^2}\,da\,dx\\&=&\int_{1}^{\pi}\frac{\pi}{2a}\,da\\&=&\color{red}{\frac{\pi}{2}\log\pi}.\end{eqnarray*}$$

Jack D'Aurizio
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  • You have to check almost nothing going this way. You write $\arctan(\pi x)-\arctan(x)$ as an integral and switch the order of integration - you are allowed to do so since $\frac{1}{1+a^2 x^2}$ is non-negative and fulfills the hypothesis of Fubini's theorem. – Jack D'Aurizio Jun 01 '15 at 02:26
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What you have done is correct till $J'(b)=\frac{\pi}{2b}$ What you need to find is $J(\pi)-J(1)$ Therefore,integrate $\frac{\pi}{2b}$ over b from 1 to $\pi$

Akshay Bodhare
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Recall Frullani's Integral

$$\int_0^{\infty}\frac{f(\alpha t)-f(\beta t)}{t}dt=\left(f(0)-f(\infty)\right)\log(\beta/\alpha) \tag 1$$

where $f$ is continuous, the integral converges, and $f(\infty)=\lim_{t\to \infty}f(t)$.

For the integral of interest we have $f=\arctan t$, $\alpha =\pi$, and $\beta =1$. We therefore have immediately from $(1)$

$$\int_0^{\infty}\frac{\arctan(\pi t)-\arctan( t)}{t}dt=\frac{\pi}{2}\log \pi$$

NOTE:

To answer the question at the end of the original post, we ought not evaluate $J$ since that integral diverges. This is so since the arctangent function is bounded as approaches $\pi /2$ as $x \to \infty$, while the integral of $1/x$ diverges.

We could form, however, a new function $K(a)=\int_0^{\infty}\frac{\arctan(ax)-\arctan(x)}{x}dx$. Differentiating with respect to $a$ gives the result that

$$K'(a)=\int_0^{\infty}\frac{1}{1+a^2x^2}dx=\frac{\pi}{2a} \tag2$$

Integrating $(2)$ yields $K(a)=\frac{\pi}{2}\log a +C$.

Now, we see that $K(1)=0 \implies C=0$ and thus we have

$$K(\pi)=\frac{\pi}{2}\log \pi$$

as expected!

Mark Viola
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    @BetterWorld You're welcome as always! But just curious as to why you refer to me as "Ma'am?" That is an inappropriate designation. I have edited my answer to address your questions. The integral is diverges because the arctangent is bounded and the integral of $1/x$ diverges at $\infty$. The property of splitting a convergent integral into two divergent ones is not permitted. I have provided a correct way forward that is similar to the one you had used, but avoids the pitfalls. Please let me know how I can improve my answer. – Mark Viola Jun 01 '15 at 21:08
  • @betterworld The integral $\int_0^{\infty}\frac1 x,dx$ diverges even though $1/x \to 0$ as $x\to \infty$. There are common basic methods to test for convergence of an integral. Consult most any calculus text book and there will likely be a chapter devoted to that topic. One such method is based on the idea of a dominated function. Here, if the integral of $f$ converges and $|f|>|g|$, then the integral of $g$ converges also. Likewise if the integral of $g$ diverges, then the integral of $f$ does also. – Mark Viola Jun 02 '15 at 14:41
  • Yes, you are correct! If $|f|>|g|$ and the integral of $f$ diverges, we cannot conclude anything about the convergence of the integral of $g$. – Mark Viola Jun 02 '15 at 16:03
  • You're welcome! It is always my pleasure. – Mark Viola Jun 02 '15 at 16:06