Tribute to Ramanujan on his upcoming birthday!

As the question is a tribute to Sir Ramanujan, the question involves two things related to him.

$1) ~ Nested ~ Radical$ equation 26

$2)~\text{Ramanujan Master Theorem}$

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$\large{x+n+a=\sqrt{ax+(n+a)^2+x \sqrt{a(x+n)+(n+a)^2+.....}}}$

It would be nice if someone provides the proof for this because i don't know :P

Here $n=a=1729$

Also

$\large{\ln(\Gamma (1+x))=- \gamma x+ \displaystyle \sum_{k=2}^{\infty}\frac{\zeta(k)}{(k)}(-x)^{k}}$

$\large{\ln(\Gamma (1+x))=- \gamma x+ \displaystyle \sum_{k=0}^{\infty}\frac{\zeta(k+2)}{(k+2)}(-x)^{k+2}}$

For Proof see Problem number 7 at the last.

$\large{ \rightarrow \frac{\ln(\Gamma (1+x))+\gamma x}{x^2}= \displaystyle \sum_{k=0}^{\infty}\frac{\Gamma(k+1) \zeta(k+2)}{(k+2)}\frac{(-x)^{k}}{k!}}$

Thus the integral now becomes

$\large{\displaystyle \int^{\infty}_{0} (x+1729+1729-3458)^{- \frac{1}{2}} \frac{\ln(\Gamma (1+x))+\gamma x}{x^2} dx}$

$\large{= \displaystyle \int^{\infty}_{0} x^{- \frac{1}{2}} \frac{\ln(\Gamma (1+x))+\gamma x}{x^2} dx}$

Now $\large{\phi(k)=\frac{\Gamma(k+1) \zeta(k+2)}{(k+2)}}$

thus by Ramanujan Master theorem

$\large{= \displaystyle \int^{\infty}_{0} x^{s-1} \frac{\ln(\Gamma (1+x))+\gamma x}{x^2} dx=\Gamma(s) \phi(-s)}$

$\large{= \displaystyle \int^{\infty}_{0} x^{s-1} \frac{\ln(\Gamma (1+x))+\gamma x}{x^2} dx=\frac{\Gamma(s) \Gamma(1-s) \zeta(2-s)}{2-s}}$

here $s=\frac{1}{2}$

Thus we have by euler reflection formula

$\Large{=\frac{\pi \zeta(3/2)}{\left( 2-\frac{1}{2} \right) \sin \left(\frac{\pi}{2} \right)}}$

$\Large{\frac{2 \pi }{3} \zeta(3/2)=\boxed{5.46}}$

I solved it. @Aditya Kumar What do you want to generalise .. What tanishq gave ..are you satisfied with that ?

The answer is $= \dfrac{2}{3}\pi \zeta{(\dfrac{3}{2})}$