# Show that $\int_0^\infty \frac{\tan^{-1} ax-\tan^{-1} x}{x}~dx = \frac{1}{2}\pi ~ \ln(a)$

How can I show that $$\int_0^\infty \frac{\tan^{-1} ax-\tan^{-1} x}{x}~dx = \frac{1}{2}\pi \ln(a)$$?

I tried by using $$\tan^{-1} ax-\tan^{-1} x = \int_1^a\frac{1}{1+(ux)^2}~du$$ and incorporate this in the original expression: $$\int_0^\infty \frac{1}{x}\left[ \int_1^a \frac{1}{1+(ux)^2}~du\right] dx$$

and then switching the order of integration: $$\int_0^a \left[~ \int_1^\infty \frac{1}{x}~\frac{1}{1+(ux)^2}~dx\right] du$$ $$\\$$

which doesn't seem to lead anywhere. But I somehow want to use $$\int_0^\infty \frac{\sin x}{x}~dx=\frac{1}{2}\pi$$, because the arctan function is involved in the process of showing this by differentiating $$\int_0^\infty e^{-ux}\frac{\sin x}{x}~dx$$ with respect to u (aka Feynman's method). The lower bond here is $$0$$ though, and in the integral above the lower bond is $$1$$, so I don't know if that's the right way to go either. Or is there possibly a different way to try here? $$\\$$ $$\\$$ (The choice of the Fourier-tags is because this is from a book on the subject)

• This is your partial answer. There probably is a trick though. Aug 2, 2021 at 20:43
• Frullani's integral with $f(x)=\tan^{-1}x$ and $b=1$ Aug 2, 2021 at 20:45
• You were very close, but you forgot the extra factor of $x$ in the derivative that cancels out the $\frac{1}{x}$ Aug 2, 2021 at 20:56

You should be able to differentiate under the integral sign giving you:

$$I'(a)= \int_{0}^{\infty}\frac{dx}{1+a^2x^2}$$

This function is extremely easy to integrate so we get:

$$I'(a) = \frac{\pi}{2a}$$

And you should be able to do the rest :) Hope this helps

I like the answers given, but your approach works too, and is arguably (conceptually) simpler.

First, a small correction $$\int \frac{1}{1 + (ux)^2} \mathrm{d}u = \frac{1}{x}\arctan(ux),$$ not just $$\arctan(ux)$$. This means that the inner integral should be multiplied with a factor of $$x$$. This, of course, is very convenient, since it will cancel with the $$1/x$$ sitting outside.

Calling the original integral $$I$$, and making the correction, we have $$I = \int_0^\infty \frac{1}{x} \int_1^a \frac{x}{1 + (ux)^2} \mathrm{d}u \mathrm{d}x.$$ Hitting this with Tonelli's theorem gives $$I = \int_{1}^a \int_0^\infty \frac{1}{1+ (ux)^2} \mathrm{d}x \mathrm{d}u = \int_1^a \left.\frac{1}{u} \arctan(ux)\right|_{0}^\infty \mathrm{d}u\\ = \int_1^a \frac{\pi/2}{u} \mathrm{d}u = \frac{\pi}{2} \ln(a).$$

Let $$f$$ be a function differentiable on $$(0,\infty)$$. Denote $$f(\infty) = \lim_{x \to \infty} f(x)$$ If $$a,b > 0,$$ (and if $$f$$ is nice enough for Fubini's theorem to apply below), then $$\int_0^\infty \frac{f(ax) -f(bx)}{x}\mathrm{d}x = \int_0^\infty \frac{1}{x} \int_b^a xf'(ux)\mathrm{d}u \mathrm{d}x = \int_b^a \int_0^\infty f'(ux) \mathrm{d}x\mathrm{d}u\\ = \int_b^a \frac{1}{u} (f(\infty) - f(0))\mathrm{d}u = (f(\infty) - f(0)) \ln\frac{a}{b}.$$

Let $$X>0$$, then substituting $$s=ax$$ gives that $$\int_0^X \frac{\tan^{-1}(ax)}{x}dx=\int_0^{aX}\frac{\tan^{-1}(s)}{s}ds$$ therefore $$\int_0^X\frac{\tan^{-1}(ax)-\tan^{-1}(x)}{x}dx=\int_X^{aX}\frac{\tan^{-1}(x)}{x}dx$$ Using the mean value theorem for integrals, there exists $$c_X\in (X,aX)$$ (or $$(aX,X)$$ if $$a<1$$) such that $$\int_X^{aX}\frac{\tan^{-1}(x)}{x}dx=\tan^{-1}(c_X)\int_X^{aX}\frac{dx}{x}=\tan^{-1}(c_X)\log(a)$$ Since $$c_X\geqslant\min(X,aX)$$, we have $$\lim\limits_{X\rightarrow +\infty}c_X=+\infty$$ and thus, letting $$X\rightarrow +\infty$$, $$\int_0^{+\infty}\frac{\tan^{-1}(ax)-\tan^{-1}(x)}{x}dx=\frac{\pi}{2}\log(a)$$

• Really interesting method! Aug 6 at 3:08

In general, if $$f$$ is differentiable on $$(0,\infty)$$ and such that the limits $$\displaystyle f(\infty):=\lim_{x\to\infty}f(x)$$ and $$\displaystyle f(0^+):=\lim_{x\to 0^+}f(x)$$ exist and are finite, then for every $$r, s >0$$ we have $$I(s,r):=\int_0^{\infty}\frac{f(sx)-f(rx)}{x}dx=(f(\infty)-f(0^+))\ln\left(\frac{s}{r}\right)$$ Proof: Setting $$u=rx$$ And $$t=s/r$$, we get $$I(s,r)=\int_0^{\infty}\frac{f(tu)-f(u)}{u}du=I(t,1)$$

Since $$t>0$$, we have

$$\begin{eqnarray} \frac{\partial I}{\partial t}(t,1) &=&\int_0^{\infty}f’(tu)du \cr &=&\frac{1}{t}\int_0^{\infty}f’(y)dy \cr &=&\frac{f(\infty)-f(0^+)}{t} \end{eqnarray}$$ By integrating we obtain $$I(t,1)=I(1,1)+(f(\infty)-f(0^+))\ln(t).$$ Because $$I(1,1)=0$$, we get $$I(s,r)=(f(\infty)-f(0^+))\ln\left(\frac{s}{r}\right).$$

Another way to go about it would be to use double integration

$$\begin{eqnarray} I(s,r) &=&\int_0^{\infty}\frac{f(sx)-f(rx)}{x}dx\cr &=&\int_0^{\infty}\int_r^s f'(yx)dydx\cr &=&\int_r^s \int_0^{\infty} f'(yx)dxdy\cr &=&\int_r^s\frac{1}{y}\left[f(yx)\right]_{x\to 0^+}^{x\to \infty} dy \cr &=&\int_r^s\frac{f(\infty)-f(0^+)}{y}dy \cr &=&(f(\infty)-f(0^+))\ln(y)\Big|_r^s \cr &=&(f(\infty)-f(0^+))(\ln(s)-\ln(r))\cr I(s,r)&=&(f(\infty)-f(0^+))\ln\left(\frac{s}{r}\right) \end{eqnarray}$$

Now, for $$f(x)=\tan^{-1}(x)$$ we have $$f(\infty)=\frac{\pi}{2}, \quad f(0^+)=0$$ It follows that $$\int_0^{\infty}\frac{\tan^{-1}(ax)-\tan^{-1}(x)}{x}dx=\frac{\pi}{2}\ln(a)$$