<s>[INST] Why does this proof fail? Catalan Constant's Exact Value. A failed attempt at finding an exact value for Catalan's Constant "C"
Definition :
$$ C+ \int_{\frac{\pi}{2}}^{\pi} \frac{x \sin{(x)}}{\sqrt{1+\cos{(x)}^2}} = \pi \ln(1+\sqrt{2})
 $$
Let
$$I(b) = \int_{\frac{\pi}{2}}^{\pi} \frac{x \sin{(bx)}}{\sqrt{1+\cos{(x)}^2}}$$
such that $ I(b=1) $ is our desired integral.
Integrate both sides with respect to b (Ad = antiderivative)
$$ I^{Ad} (b)= -\int_{\frac{\pi}{2}}^{\pi} \frac{\cos{(bx)}}{\sqrt{1+\cos{(x)}^2}} $$
I shall restrict b to be an odd integer greater than $0$
I Examine the values given by the integral :
$$ p = 1 \to \left (\frac{\pi}{4}\right)$$
$$ p = 3 \to \left(2-\frac{3 \pi}{4}\right) $$
$$ p = 5  \to \left(\frac{13 \pi }{4} - 10\right) $$
I divide these values into Pi Numbers ( those with $\pi$) and Rational Numbers (those without it).
The Pi Numbers will follow a known sequence called The Central Delannoy Numbers $D(n)$
$$ D(n) = \sum_{k=0}^{n} \binom{n}{k}^{2} 2^k $$
Putting this into b terms .
$$ \Pi (b) = \frac{\pi}{4}  \sum_{k=0}^{\frac{b-1}{2}}\binom{\frac{b-1}{2}}{k}^2\, 2^k $$
For the Rational Numbers $R(b)$ we notice that since b is an integer then the $\cos(bx)$ may be expanded with Chebyschev Polynomials in the form :
$$ \cos(bx) = b \sum_{k=0}^{b} (-2)^{k} \frac{(b+k-1)!}{(b+k)!(2k)!} (1-\cos(x))^{k} $$
Since
$$ -\int_{\frac{\pi}{2}}^{\pi} \frac{\cos(x)^{b}}{\sqrt{1+\cos(x)^2}} = \frac{\sqrt{\pi}}{4} \frac{\Gamma(\frac{b}{4}+\frac{1}{4})}{\Gamma(\frac{b}{4}+\frac{3}{4})}$$
for odd integers greater than $0$
We can give $R(b)$ as
$$ \frac{\sqrt{\pi} b}{4} \sum_{k=1}^{\lfloor \frac{b+1}{4} \rfloor} \frac{\Gamma(k)}{\Gamma(k+\frac{1}{2})} \frac{4^{2k-1}}{4k-1}\binom{\frac{b+4k-3}{2}}{{4k-2}} $$
Together :
$$ -\int_{\frac{\pi}{2}}^{\pi} \frac{\cos{(bx)}}{\sqrt{1+\cos{(x)}^2}} = \sqrt{2}  \cos\left(\frac{\pi}{4}-\frac{\pi b}{2}\right)\left(\frac{\pi}{4}  \sum_{k=0}^{\frac{b-1}{4}}\binom{\frac{b-1}{2}}{k}^2\, 2^k - \frac{\sqrt{\pi} b}{4} \sum_{k=1}^{\lfloor \frac{b+1}{4} \rfloor} \frac{\Gamma(k)}{\Gamma(k+\frac{1}{2})} \frac{4^{2k-1}}{4k-1}\binom{\frac{b+4k-3}{2}}{{4k-2}}\right)$$
All that is left is to take the derivative with respect to b and let $b = 1$ ( I am unsure of the constant that is generated when we integrate with respect to b  , don't know if it goes to $0$ after we take the derivative again )
I do not know how to take the derivative of b in its current form so i decided to change it into integral form.
The Central Delannoy Numbers can be given with :
$$ D(n) = \frac{1}{\pi} \int_{3-2\sqrt{2}}^{3+2\sqrt{2}} \frac{1}{\sqrt{-t^2+6t-1}} \frac{1}{t^{n+1}} \,dt $$
The generating function for $R(b) $ is
$$ \frac{\sin^{-1}\left(\frac{4x}{(1-x)^2}\right)}{2\sqrt{x^2-6x+1}} $$
Putting it all together :
$$ -\int_{\frac{\pi}{2}}^{\pi} \frac{\cos{(bx)}}{\sqrt{1+\cos{(x)}^2}}= \cos\left(\frac{\pi}{4}-\frac{\pi b}{2}\right) \left(\frac{1}{2 \sqrt{2}} \int_{3-2\sqrt{2}}^{3+2\sqrt{2}} \frac{t^{-b-\frac{1}{2}}}{\sqrt{-t^2+6t-1}} \, dt -\sqrt{2} \,\lim_ {t \to 1} \int \frac{t^{-b-\frac{1}{2}}}{\sqrt{-t^2+6t-1}}\, dt \right) $$
Unfortunately, taking the derivative with respect to b and letting $b=1$ yields
$$ \frac{1}{2} \lim_{t \to 1} \int \frac{\ln{t}}{t} \frac{1}{\sqrt{-t^2+6t-1}} \,dt $$
which cancels the Catalan Constant on the LHS.
Question(s) : Does anybody see another way to proceed which could possibly lead to better results?
Does anyone see any mistakes?
Thank you kindly for your help and time.
 [/INST] Integrate by parts
\begin{align}
\int_{\frac{\pi}{2}}^{\pi} \frac{x \sin{x}}{\sqrt{1+\cos^2{x}}} dx
=-
 \int_{\frac{\pi}{2}}^{\pi} x d(\sinh^{-1}\cos{x})
=\pi \ln(1+\sqrt{2}) - J(1)
\end{align}
where $J(a)= \int^{\frac{\pi}{2}}_{0} \sinh^{-1}(a\cos t) dt$. Evaluate
$$
J’(a)= \int^{\frac{\pi}{2}}_{0} \frac{\cos{t}}{\sqrt{1+a^2\cos^2{t}}} dt
=\frac1a \sin^{-1}\frac a{\sqrt{1+a^2}}= \frac{\tan^{-1}a}a
$$
$$J(1)= \int_0^1 \frac{\tan^{-1}a}ada=-\int_0^1\frac{\ln a}{1+a^2}da=C
$$
Thus
$$ C+ \int_{\frac{\pi}{2}}^{\pi} \frac{x \sin{x}}{\sqrt{1+\cos^2x}} = \pi \ln(1+\sqrt{2})
 $$
</s>