Another integral representation of Catalan's Constant?

Question

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. :)

Answer 1

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}

Answer 2

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.