Consider $\displaystyle \int_{0}^{\infty}\frac{\log(x)\cot^{-1}(x)}{\sqrt{x}}$.

I've tried: $\displaystyle F(a,b) = \int_{0}^{\infty} \frac{\log(ax)\cot^{-1}(bx)}{\sqrt{x}}$, so $\displaystyle F''(a,b) = \int_{0}^{\infty}\frac{dx}{a\sqrt{x}(1+b^2x^2)} = \int_{0}^{\infty}\frac{dt}{2a(1+x^4b^2)} = \frac{\pi}{2a\sqrt{2b}}$.

So $F'_{a}(a,b) = \frac{\pi}{2a}\sqrt{b}+C(a)$, also we can make for $a$. But how can we find constant?

It's easy to see that $C(a) = 0$, what about $C(b)$? If we consider $F'_{b}(a,b) = \frac{\pi\log(a)}{\sqrt{8b}} + C(b)$ then it isn't easy to find it. Any ideas?

edit also I thought about consider $\cot^{-1}(bx)$ and then make a substitution $t = \frac{1}{1+x}$ and represent $\log$ as series

  • 1
    $\begingroup$ It seems your integral does not converge. $\endgroup$
    – omegadot
    Dec 24 '17 at 7:39
  • $\begingroup$ @omegadot yeah there must be $arcctg$. But it's doesn't help $\endgroup$
    – openspace
    Dec 24 '17 at 7:41
  • $\begingroup$ "arcctg" is not a standard mathematical notation for a trigonometric function. Use standard notation. Either it is $\operatorname{arccot}$ or $\cot^{-1}$ or $\arctan$ or $\tan^{-1}$. $\endgroup$
    – heropup
    Dec 24 '17 at 9:13
  • $\begingroup$ @heropup sorry in my language it oftener write like $arcctg$ $\endgroup$
    – openspace
    Dec 24 '17 at 9:25
  • $\begingroup$ @heropup but I think , that this's not a mistake. Users understand what the main idea. $\endgroup$
    – openspace
    Dec 24 '17 at 9:27

Here is a slightly different approach: Consider

$$ I(s) = \int_{0}^{\infty} \frac{\arctan x}{x^{1+s}} \, dx. $$

This integral converges when $s \in (0, 1)$. And for this $s$, taking integration by parts followed by the substitution $x = \tan\theta$ leads to

$$ I(s) = \frac{1}{s}\int_{0}^{\infty} \frac{1}{x^s (1+x^2)} \, dx = \frac{1}{2s}\operatorname{B}\left(\frac{1-s}{2},\frac{1+s}{2}\right) = \frac{\pi}{2s}\sec\left(\frac{\pi s}{2}\right). $$

On the other hand, the substitution $x \mapsto 1/x$ shows that our integral is written as

$$ \int_{0}^{\infty} \frac{\log x \operatorname{arccot}(x)}{x^{1/2}} \, dx = - \int_{0}^{\infty} \frac{\log x \arctan x}{x^{3/2}} \, dx = I'(1/2). $$

Evaluating the derivative, we finally obtain

$$ \int_{0}^{\infty} \frac{\log x \operatorname{arccot}(x)}{x^{1/2}} \, dx = \frac{\pi^2}{\sqrt{2}} - 2\sqrt{2} \, \pi. $$


It seems that the integral does not converge on $[1, \infty]$ for example, therefore not on $[0, \infty]$:

$\lim_{x \to \infty} \frac {\frac {\log(x)\arctan(x)} {\sqrt x}} {\frac 1 {\sqrt x}} = \lim_{x \to \infty} \log(x) \arctan(x) = \infty$

and since $\int_1^\infty \frac {dx} {\sqrt x}$ diverges, your integral diverges by the second comparison test.

  • $\begingroup$ First of all there is $arcctg(x)$. $\endgroup$
    – openspace
    Dec 24 '17 at 7:59
  • 1
    $\begingroup$ Correct... I don't see your point $\endgroup$
    – AsafHaas
    Dec 24 '17 at 8:06

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