3

I am trying to find closed form for this integral:

$$I(a)=\int_0^1\frac{\arctan{ax}}{\sqrt{1-x^{2}}}dx$$

Where $a>0$.

My try: Let: $$I(a)=\int_0^1\frac{\arctan{ax}}{\sqrt{1-x^{2}}}dx$$

Then: $$\frac{dI(a)}{da}=\int_0^1\frac{x}{(1+(ax)^2)\sqrt {1-x^2}}dx$$

How I can complete this work or is there another way to approach it?

Zacky
  • 27,674
  • Have you tried substitution $x=\sin u$, standard for such integrals? – Adam Latosiński Apr 29 '19 at 23:41
  • No , ever but I think make integral so difficult –  Apr 29 '19 at 23:51
  • 2
    I've tried this substitution, and it lets you calculate $\frac{dI(a)}{da}$. However, integrating the resulting expression is more complicated. It doesn't seem that $I(a)$ can be expressed with simple formulas, for general $a$ you need polylogarithms to express it. – Adam Latosiński Apr 29 '19 at 23:55
  • You can use the substitution $x = \sqrt{1-u^2}$ to get $I'(a) = \ln(a+\sqrt{1+a^2})/(a\sqrt{1+a^2})$ – eyeballfrog Apr 30 '19 at 00:27

3 Answers3

7

The integral can be expressed in terms of Legendre chi function: $$I(a)=\int_0^1\frac{\arctan\left(ax\right)}{\sqrt{1-x^{2}}}dx\overset{x=\sin \theta}=\int_0^\frac{\pi}{2}\arctan(a\sin \theta)d\theta =2\chi_2\left(\frac{\sqrt{1+a^2}-1}{a}\right)$$

For the first case you asked (before the edit) we can take a special value from here (see $8$): $$\chi_2\left({\sqrt{5}-2}\right)=\frac{\pi^2}{24}-\frac34 \ln^2 (\phi), \quad \phi=\frac{1+\sqrt 5}{2}$$

So for $a=\frac12$ we have: $$I=\int_0^1 \frac{\arctan\left(\frac{x}{2}\right)}{\sqrt{1-x^2}}dx=2\chi_2\left({\sqrt{5}-2}\right)=\frac{\pi^2}{12}-\frac32 \ln^2 (\phi)$$

Zacky
  • 27,674
4

A series representation for the integral.

For $|x|\leq1$ we have that $$\arctan x=\sum_{n\geq0}\frac{(-1)^n}{2n+1}x^{2n+1}$$ So Your integral is $$I(a)=\int_0^1\sum_{n\geq0}\frac{(-1)^n a^{2n+1}}{2n+1}\frac{x^{2n+1}dx}{\sqrt{1-x^2}}\\ =\sum_{n\geq0}\frac{(-1)^n a^{2n+1}}{2n+1}\int_0^1\frac{x^{2n+1}dx}{\sqrt{1-x^2}}$$ Then we focus on $$\int_0^1\frac{x^{2n+1}dx}{\sqrt{1-x^2}}\overset{x=\sin(t)}=\int_0^{\pi/2}\sin(t)^{2n+1}dt=\frac{2n+1}{2^n}\cdot\frac{(2n)!}{n!}$$ Which is similar to the integral here. So we have $$I(a)=\sum_{n\geq0}\frac{(-1)^n(2n)!}{2^n n!}a^{2n+1}=a\sum_{n\geq0}\frac{(2n)!}{n!}\left(-\frac{a^2}{2}\right)^n$$ Then we set $$t_n=\frac{(2n)!}{n!}\left(-\frac{a^2}{2}\right)^n$$ We have that $$\frac{t_{n+1}}{t_n}=(n+1)(n+1/2)\frac{-2a^2}{n+1}$$ and $t_0=1$. Hence we have the hypergeometric series representation $$I(a)=a\cdot{{}_{2}F_{0}}\left({1,\frac12 \atop -}\bigg|-2a^2\right)$$

clathratus
  • 17,161
4

Assuming $a$ to be real and positive, a CAS gives

$$\color{blue}{I(a)=\frac{\pi ^2}{4}-2 \sinh ^{-1}(a) \,\coth ^{-1}\left(a+\sqrt{a^2+1}\right)+\text{Li}_2\left(a-\sqrt{a^2+1}\right)-\text{Li}_2\left(\frac{1}{a+\sqrt{a^2+1}} \right)}$$ $$I\left(\frac{1}{2}\right)=\frac{\pi ^2}{12}-\frac{1}{2} \sinh ^{-1}(2) \,\text{csch}^{-1}(2)$$

Using @clathratus's approach $$I(a)=\sum_{n\geq0}\frac{(-1)^n a^{2n+1}}{2n+1}\int_0^1\frac{x^{2n+1}}{\sqrt{1-x^2}}dx$$ we have $$\int_0^1\frac{x^{2n+1}}{\sqrt{1-x^2}}dx=\frac{\sqrt{\pi }\, \Gamma (n+1)}{2\, \Gamma \left(n+\frac{3}{2}\right)}$$ making $$I(a)=a \, _3F_2\left(\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2};-a^2\right)$$ the simplication of which giving the first result.

Claude Leibovici
  • 260,315