# Show that $\int_0^\infty\frac{\arctan(\mathrm{πx})-\arctan(\mathrm x)}xdx=\;\frac{\mathrm\pi}2\ln(\pi)$ using 12th grade calculus.

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..

• My first guess is Feynman's Trick...? Apr 7 '18 at 17:34
• 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/… Apr 7 '18 at 17:35
• There are some more methods here, if you're interested: math.stackexchange.com/questions/460307/… Apr 7 '18 at 17:48
• 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 $\log$. Apr 7 '18 at 18:07
• 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. 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, $I(a) = \frac{\pi}{2}\ln(a)$.

• Really nice, sir. Apr 7 '18 at 17:36
• 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. Apr 7 '18 at 17:37
• @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. Apr 7 '18 at 17:38
• 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 Apr 7 '18 at 17:41
• @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. 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 $\frac{\pi}{2}\log(\pi)$.