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I tried to solve this integral on Brilliant earlier using an identity for improper integrals, but it turned out to be wrong. I did some research and found that this identity only works when our function has a "bounded anti-derivative" on our interval. Here is the integral in question:

$$\int_0^\infty \frac1x \arctan\left(\frac{3x^2}{4x^4+5x^2+2}\right) \mathrm dx$$

Here is where I found this identity: https://brilliant.org/wiki/integration-tricks/

Scroll down to where it says "Inversions" to find it.

The question I had was why you had to have a bounded anti-derivative in order to make the substitution? I understand that the answer I received from doing this doesn't make sense (as I got the integral diverges), but I don't understand why. Could anyone shed some light on this?

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  • $\begingroup$ I think the inversion would not affect convergence. If the original integral diverges then the integral after inversion diverges as well. Similarly, if the original integral converges then the integral still converges after inversion. $\endgroup$ – Szeto Sep 25 '18 at 4:06
  • $\begingroup$ I have hard time understanding what is the connection between the integral $$\int_0^\infty \frac1x \arctan\left(\frac{3x^2}{4x^4+5x^2+2}\right) dx$$ and the trick "inversion", since I dont see the integral anywhere on that page. Do you wish to solve the integral? $\endgroup$ – Zacky Sep 25 '18 at 10:26
  • $\begingroup$ Yes, I wanted to solve the integral so I used the inversion method. I guess I probably did a wrong step in my work. From what I remember, I applied that rule and then used arctan(1/x) = pi/2 - arctan(x) when x>0. Maybe I couldn't use this here, as I then had the arctans cancel out. $\endgroup$ – Zach Sep 25 '18 at 10:50
  • $\begingroup$ Do you wish to see an approach without that substitution? $\endgroup$ – Zacky Sep 25 '18 at 10:51
  • $\begingroup$ Sure, but could you try to approach it using the substitution as well and show if/why it doesn't work. Maybe my mistake was using the arctan(x) + arctan(1/x) = pi/2 $\endgroup$ – Zach Sep 25 '18 at 10:55
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$$\frac{3x^2}{1\color{blue}{+}4x^4+5x^2+1}=\frac{4x^2\color{blue}{-}x^2}{1\color{blue}{+}(4x^2+1)(x^2+1)}=\frac{(4x^2+1)\color{blue}{-}(x^2+1)}{1\color{blue}{+}(4x^2+1)(x^2+1)}$$ $$\arctan\left(\frac{(4x^2+1)\color{red}{-}(x^2+1)}{1\color{red}{+}(4x^2+1)(x^2+1)}\right)=\arctan(4x^2+1)\color{red}{-}\arctan(x^2+1)$$ Now you might notice why there was not needed any $\frac1x$ substitution. Also your integral becomes a Frullani-type integral, so: $$\int_0^\infty \frac{\arctan(4x^2+1)-\arctan(x^2+1)}{x}dx=\left(\arctan (0) -\arctan (\infty ) \right)\ln \left(\frac12 \right) =\frac{\pi}{2}\ln 2$$

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