3
$\begingroup$

I've been trying to evaluate the integral for a while now, and I've been unable to find it anywhere... I tried substituting $\tan^{-1}(tx)$ as $u$ but got nowhere... I have done dozens of other substitutions... I've been told to use the properties of gaussian integrals and the gamma function, but can't seem to figure out a way to bring in $\mathrm{e}$... A few hints that point in the right direction to think will really help!

It would be really really helpful if you could solve this using only the properties of gaussian and gamma functions!

$\endgroup$
  • 1
    $\begingroup$ Call this integral $f(t)$. Evaluate $f^\prime (t)$ with partial fractions, then use $f(0)=0$. $\endgroup$ – J.G. Mar 20 at 22:06
  • 1
    $\begingroup$ Thank you, I understand exactly what you are talking about... $\endgroup$ – Pratik Apshinge Mar 20 at 22:08
  • $\begingroup$ How will the derivative of the function give me information about the function itself? $\endgroup$ – Pratik Apshinge Mar 20 at 22:17
  • 1
    $\begingroup$ Maybe you can integrate the derivative of the function. $\endgroup$ – Minus One-Twelfth Mar 20 at 22:17
  • $\begingroup$ And, also, is there any possible way to do this with gaussian integrals or the gamma function? $\endgroup$ – Pratik Apshinge Mar 20 at 22:18
4
$\begingroup$

Let: $$ I(t)=\int_0^\infty\frac{\arctan tx}{x(1+x^2)}dx. $$ Then $$ I'(t)=\int_0^\infty\frac{1}{(1+t^2x^2)(1+x^2)}dx= \frac1{t^2-1}\int_0^\infty\left[\frac{t^2}{1+t^2x^2}-\frac{1}{1+x^2}\right]dx\\ =\frac1{t^2-1} \left[t\arctan(tx)-\arctan x\right]_0^\infty=\frac\pi2\frac{|t|-1}{t^2-1}=\frac\pi2\frac1{|t|+1}. $$ Finally integrating the last expression over $t$ one obtains: $$ I(t)=I(0)+\frac\pi2\text {sgn}(t)\log(|t|+1)=\text {sgn}(t)\frac\pi2\log(|t|+1). $$

$\endgroup$
  • $\begingroup$ Wouldn't $\left[t\tan^{-1}\left(tx\right)-\tan^{-1}x\right]^{ }$ from 0 to infinity be 0? $\endgroup$ – Pratik Apshinge Mar 20 at 22:51
  • 1
    $\begingroup$ @PratikApshinge No. It will be $(t-1)\pi/2$. $\endgroup$ – user Mar 20 at 22:54
  • $\begingroup$ I'm so sorry, made a calculation mistake! $\endgroup$ – Pratik Apshinge Mar 20 at 22:54
  • $\begingroup$ No problem. You're welcome. $\endgroup$ – user Mar 20 at 22:56
3
$\begingroup$

We can write $\arctan(tx)=\int_0^t \frac{x}{1+y^2x^2}\,dy$. Proceeding, we find that for $t>0$

$$\begin{align} \int_0^\infty \frac{\arctan(tx)}{x(1+x^2)}\,dx&=\int_0^\infty \frac{1}{1+x^2}\int_0^t \frac1{1+y^2x^2}\,dy\,dx\\\\ &\overbrace{=}^{\text{Fubini}}\int_0^t \int_0^\infty \frac{1}{(1+x^2)(1+y^2x^2)}\,dx\,dy\\\\ &=\int_0^t \frac{\pi/2}{y+1}\,dy\\\\ &=\frac\pi2 \log(1+t) \end{align}$$

Inasmuch as the integral of interest is an odd function of $t$, we have

$$\int_0^\infty \frac{\arctan(tx)}{x(1+x^2)}\,dx=\begin{cases}\frac\pi 2\log(1+t)&, t>0\\\\0&,t=0\\\\-\frac\pi2 \log(1-t)&,t<0\end{cases}$$

$\endgroup$

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.