8

The integral in question: $$I=\int_{0}^{1} {\frac{\arctan\left(\frac{x-1}{\sqrt{x^2-1}}\right)}{\sqrt{x^2-1}}}dx.$$

So I have a solution and it's rather straightforward, but I don't much like it. I'm not sure how to tackle this integral head-on, as you'll see, and I was wondering if anyone had another way of arriving at the result.

My Solution:

Let

$$I(a)=\int_{0}^{\frac{\pi}{2}} {\ln{(\sin{(x)}+a)}}\hspace{1pt}dx.$$

Then

$$I'(a)=\int_{0}^{\frac{\pi}{2}} {\frac{1}{\sin{(x)}+a}}dx=\frac{2}{\sqrt{a^2-1}}\arctan{\left(\frac{a-1}{\sqrt{a^2-1}}\right)}.$$

To show this one can apply the Weierstrass Substitution and complete the square in the denominator.

Now consider the following

$$\int_{0}^{1} {\frac{2}{\sqrt{x^2-1}}\arctan{\left(\frac{x-1}{\sqrt{x^2-1}}\right)}}dx=I(1)-I(0).$$

Noticing the famous result $I(0)=-{\pi}\ln(2)/2$ (check 1 for more details) we have

$$I=\frac{\pi}{4}\ln{(2)}+\frac{1}{2}\int_{0}^{1} {\ln{(\sin{(x)}+1)}}\hspace{1pt}dx.$$

Using

$$\sin{(x)}+1=2\sin{\left(\frac{1}{2}\left(x+\frac{\pi}{2}\right)\right)}\cos{\left(\frac{1}{2}\left(x-\frac{\pi}{2}\right)\right)}$$

we arrive at the following $$I=\frac{\pi}{2}\ln{(2)}+2\int_{0}^{\frac{\pi}{4}} {\ln{(\cos{(x)})}}\hspace{1pt}dx$$

and now by 2 we have $I=C$, where $C$ is Catalan's Constant.

If you have seen this integral before please let me know where and again if you have a different solution that would be much appreciated. :)

Alejandro Jimenez Tellado
  • 290
  • 1
  • 8
  • 1
    Essentially equivalent to identities $(7)$ and $(8)$ here if you retrace your steps. But cool nonetheless, +1. – K.defaoite Jul 07 '23 at 00:23
  • @K.defaoite I suppose so, though I really want to approach the integral all on its own, as if you had seen it out on its own and had no idea of the connection with Catalan's Constant. Of course, thank you for the upvote. – Alejandro Jimenez Tellado Jul 07 '23 at 00:28
  • 2
    @Gonçalo Surprisingly no, in fact, the integral of that from 0 to 1 is $\frac{{\pi}^2}{16}$. This isn't too hard to see as that integral actually has a nice antiderivative. – Alejandro Jimenez Tellado Jul 07 '23 at 01:04

2 Answers2

12

Note that \begin{align} \int_{0}^{1} {\frac{\tan^{-1}\frac{x-1}{\sqrt{x^2-1}}}{\sqrt{x^2-1}}}dx =& \int_{0}^{1} {\frac{\tanh^{-1}\sqrt{\frac{1-x}{1+x}}}{\sqrt{1-x^2}}}\overset{y=\sqrt{\frac{1-x}{1+x}} }{dx}\\ =&\ 2\int_0^1\frac{\tanh y}{1+y^2}\overset{y\to \frac{1-y}{1+y} }{dy}= -\int_0^1 \frac{\ln y}{1+y^2}dy=C \end{align}

Quanto
  • 97,352
7

Using the identity $\arctan(iz)=i\tanh^{-1}(z)$, we may rewrite the integral as $$ I=\int_{0}^{1} {\frac{\tanh^{-1}\left(\frac{1-x}{\sqrt{1-x^2}}\right)}{\sqrt{1-x^2}}}\,dx. \tag{1} $$ Substituting $x$ with $\text{sech}\,u$, we may rewrite $(1)$ as \begin{align} I&=\int_{\infty}^{0} \frac{\tanh^{-1}\left(\tanh(u/2)\right)}{\tanh u}\,(-\tanh u\,\text{sech}\,u)\,du \\ &=\frac{1}{2}\int_0^{\infty}\frac{u}{\cosh u}\,du, \tag{2} \end{align} which is integral (5) in the paper linked by @K.defaoite.

Gonçalo
  • 9,312