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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)

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

4 Answers 4

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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

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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 $$ I = \int_{1}^a \int_0^\infty \frac{1}{1+ (ux)^2} \mathrm{d}x \mathrm{d}u = \int_1^a \left.\frac{1}{u} \arctan(ux)\right|_{0}^\infty \mathrm{d}u\\ = \int_1^a \frac{\pi/2}{u} \mathrm{d}u = \frac{\pi}{2} \ln(a).$$


One nice thing about this argument is that it generalises completely to the Frullani integral mentioned in the comments.

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 $$ \int_0^\infty \frac{f(ax) -f(bx)}{x}\mathrm{d}x = \int_0^\infty \frac{1}{x} \int_b^a xf'(ux)\mathrm{d}u \mathrm{d}x = \int_b^a \int_0^\infty f'(ux) \mathrm{d}x\mathrm{d}u\\ = \int_b^a \frac{1}{u} (f(\infty) - f(0))\mathrm{d}u = (f(\infty) - f(0)) \ln\frac{a}{b}. $$

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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 $X\rightarrow +\infty$, $$ \int_0^{+\infty}\frac{\tan^{-1}(ax)-\tan^{-1}(x)}{x}dx=\frac{\pi}{2}\log(a) $$

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  • $\begingroup$ Really interesting method! $\endgroup$
    – K.defaoite
    Aug 6 at 3:08
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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 $$ \int_0^{\infty}\frac{\tan^{-1}(ax)-\tan^{-1}(x)}{x}dx=\frac{\pi}{2}\ln(a) $$

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