I have to show that $\int_0^\infty\frac{\arctan(\mathrm{πx})-\arctan(\mathrm x)}xdx=\;\frac{\mathrm\pi}2\ln(\pi)$ using 12th grade calculus wich means single variable calculus.

What I've tried:

First I tried to make a single arctan from those 2: $\arctan(\mathrm{πx})-\arctan(\mathrm x)=\arctan(\frac{(\mathrm\pi-1)\mathrm x}{1+\mathrm{πx}^2})$ as you can see it's still not very pleasant... and it looks like I don't get anywhere with this..

  • 3
    $\begingroup$ My first guess is Feynman's Trick...? $\endgroup$ – Crescendo Apr 7 '18 at 17:34
  • $\begingroup$ The general result is derived here using single variable calculus but usually Leibniz's rule won't have been taught. math.stackexchange.com/questions/1021468/… $\endgroup$ – Joshua Farrell Apr 7 '18 at 17:35
  • $\begingroup$ There are some more methods here, if you're interested: math.stackexchange.com/questions/460307/… $\endgroup$ – Jonathan Apr 7 '18 at 17:48
  • $\begingroup$ For this particular problem, usual tricks (differentation under the integral sign, Frullani's theorem, Laplace transform etc) can be emulated by simple substitutions and symmetry, due the "implicit" relation between $\arctan$ and $\log$. $\endgroup$ – Jack D'Aurizio Apr 7 '18 at 18:07
  • 1
    $\begingroup$ Actually my elementary solution exploits a very deep idea, i.e. that every hypergeometric function of the $\phantom{}_2 F_1$ kind fulfills some kind of peculiar symmetry relation. $\endgroup$ – Jack D'Aurizio Apr 7 '18 at 18:09

Let $I(a) = \int_0^\infty \frac{\arctan(a x)-\arctan(x)}{x}dx$. Then $I'(a) = \int_0^\infty \frac{1}{x}\frac{x}{1+(ax)^2}dx = \frac{1}{a}\frac{\pi}{2}$. So, $I(a) = \frac{\pi}{2}\ln(a)+C$. Letting $a=1$, we see that $0=I(1)=\frac{\pi}{2}\ln(1)+C \implies C=0$. So, $I(a) = \frac{\pi}{2}\ln(a)$.

  • 1
    $\begingroup$ Really nice, sir. $\endgroup$ – C. Cristi Apr 7 '18 at 17:36
  • 2
    $\begingroup$ That is fine if the OP is aware of the theorem of differentiation under the integral sign, which might not be a topic in 12th grade Calculus. $\endgroup$ – Jack D'Aurizio Apr 7 '18 at 17:37
  • $\begingroup$ @JackD'Aurizio It's not really a topic in 12th grade Calculus but I'm really intrested in integrals and integration questions/techniques and I am aware of the Feynman's trick and it's existence but nobody really taught me how to use it... I just know some little things about it but I am very very familiar with other techniques of integration. $\endgroup$ – C. Cristi Apr 7 '18 at 17:38
  • 2
    $\begingroup$ I personally view it as a 12th grade calculus. Nobody usually cares about justifying these sorts of tricks at that point. Some things I really don't consider 12th grade calculus are multivariate tricks, complex analytic tricks, fourier/legendre transform, etc $\endgroup$ – mathworker21 Apr 7 '18 at 17:41
  • 2
    $\begingroup$ @C.Cristi It depends. Some substitutions make sense, like getting rid of a natural log in the denominator, while other times it seems completely arbitrary. In this case, I would probably go along the lines of thinking that differentiating w.r.t a gives an x in the numerator, which we can cancel out in the denominator. $\endgroup$ – Crescendo Apr 7 '18 at 17:50

As pointed out this can be solved through differentiation under the integral sign (Feynman's trick), or simply invoking Frullani's theorem. Since I doubt they are topics in 12th grade Calculus, I will try to outline a single-variable approach as elementary as possible. By integration by parts $$ \int_{0}^{+\infty}\frac{\arctan(\pi x)-\arctan(x)}{x}\,dx = \int_{0}^{+\infty}\left(\frac{1}{1+x^2}-\frac{\pi}{1+\pi^2 x^2}\right)\log(x)\,dx $$ and by the substitution $x=e^t$ the RHS turns into $$ \int_{-\infty}^{+\infty}\left(\frac{1}{1+e^{2t}}-\frac{\pi}{1+\pi^2 e^{2t}}\right) te^t\,dt=\pi\int_{0}^{+\infty}\left(\frac{1}{\pi^2+ e^{2t}}-\frac{1}{1+\pi^2 e^{2t}}\right)te^t\,dt $$ since $\int_{-\infty}^{+\infty}g(t)\,dt = \int_{0}^{+\infty}\left(g(t)+g(-t)\right)\,dt$ as soon as $g\in L^1(\mathbb{R})$. Now we may break the integral into two convergent integrals (we cannot do this at the very beginning since $\frac{\arctan x}{x}$ behaves like $\frac{C}{x}\not\in L^1(1,+\infty)$ for large values of $x$) $$\mathcal{J}_1=\int_{0}^{+\infty}\frac{t e^t}{\pi^2+e^{2t}}\,dt = \int_{1}^{+\infty}\frac{\log(u)}{\pi^2+u^2}\,du$$ $$\mathcal{J}_2=\int_{0}^{+\infty}\frac{t e^t}{1+\pi^2 e^{2t}}\,dt = \int_{1}^{+\infty}\frac{\log(u)}{1+\pi^2 u^2}\,du=\int_{0}^{1}\frac{-\log(u)}{\pi^2+u^2}\,du$$ then re-combine them into $$\mathcal{J}_1-\mathcal{J}_2 = \int_{0}^{+\infty}\frac{\log(u)}{\pi^2+u^2}\,du\stackrel{u\mapsto \pi v}{=}\frac{1}{\pi}\int_{0}^{+\infty}\frac{\log\pi+\log v}{1+v^2}\,dv. $$ Last trick: $\int_{0}^{+\infty}\frac{\log v}{1+v^2}\,dv$ equals zero by $v\mapsto e^t$. This leads to: $$ \mathcal{J}_1-\mathcal{J}_2 = \frac{\log \pi}{\pi}\int_{0}^{+\infty}\frac{dv}{1+v^2}=\frac{\log\pi}{2} $$ and to the claim $$ \int_{0}^{+\infty}\frac{\arctan(\pi x)-\arctan(x)}{x}\,dx = \color{red}{\frac{\pi\log\pi}{2}}.$$


While not really 12th grade level, Glaisher's theorem should be of interest to the OP, given the interest in learning about integration methods. Glaisher's theorem is a special case of Ramanujan's master theorem, it says:

If $f(x)$ is an even function with series expansion around $x = 0$:

$$f(x)= \sum_{k=0}^{\infty}(-1)^kc_k x^{2k}$$

and the integral over the real line converges, then we have:

$$\int_{0}^{\infty}f(x) dx = \frac{\pi}{2} c_{-\frac{1}{2}}$$

$c_n$ is a priori only defined for integer $n$, but when an analytic expression for $c_n$ is known then one should be able to substitute $n = -\frac{1}{2}$. Here expressions involving factorials should be replaced by gamma functions, often one deals with simple rational functions in which case putting $n = -\frac{1}{2}$ poses no problems. In some cases one needs to take the limit $n \rightarrow -\frac{1}{2}$.

In this case, we have:

$$c_n = \frac{\pi^{2n+1} - 1}{2 n + 1}$$

for $n = -\frac{1}{2}$ this becomes ill defined, but as mentioned above, we then need to take the limit for $n \rightarrow -\frac{1}{2}$. This yields:

$$\lim_{n \rightarrow -\frac{1}{2}}c_n = \log(\pi)$$

The integral thus equals $\frac{\pi}{2}\log(\pi)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.