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Start with,

$\displaystyle \sum_{k \ge 2} \zeta(k)(-x)^k = \sum_{k \ge 2}\sum_{s\ge 1} \frac{(-1)^k x^k}{s^k}$

Interchanging summations we have ,

$\displaystyle \sum_{s \ge 1} \sum_{k \ge 2} \frac{(-x)^k}{s^k} = \sum_{s \ge 1} \frac{\frac{x^2}{s^2}}{1+\frac{x}{s}} = \sum_{s \ge 1} \frac{x^2}{s(s+x)}$

So, $\displaystyle \sum_{k \ge 2} \zeta(k)(-x)^{k} = x\sum_{s\ge 1} (\frac{1}{s} - \frac{1}{s+x}) = -x H_x$

We know , $\displaystyle H_x = \psi(1+x) + \gamma$

Substituting we have , $\displaystyle \sum_{k \ge 2} \zeta(k)(-x)^{k} = x(\psi(1+x)+\gamma)$

Now changing $x \to -x$ & dividing by $x$ we have ,

$\displaystyle \sum_{k=2}^{\infty} \zeta(k)x^{k-1} = -\psi(1-x) -\gamma$

With $-x\to x$ we have , $\displaystyle \gamma+\sum_{k=2}^{\infty} \zeta(k)(-x)^{k-1} = -\psi(1+x)=-d(\ln\gamma(1+x))$ and integrating both sides from $0$ to $x$ we get,

$\displaystyle \gamma x-\sum_{k=2}^{\infty}\frac{\zeta(k)}{k}(-x)^k =-\ln\Gamma(1+x)$ and thus $\displaystyle \frac{\gamma x+\ln\gamma(1+x)}{x^2}=\sum_{k=0}^{\infty}\frac{\zeta(k+2)}{k+2}(-x)^k \times \frac{\color{#D61F06}{\Gamma(1+k)}}{\color{#D61F06}{\Gamma(1+k)}}$

Applying Ramanujan's Master Theorem we have,

$\displaystyle \int_{0}^{\infty} x^{s-3}(\gamma x+\ln\Gamma(1+x)) dx = \Gamma(s)\frac{\zeta(2-s)}{2-s}\Gamma(1-s)$

Putting $s=\frac{3}{4}$ we have $\displaystyle \int_{0}^{\infty} \frac{\ln\Gamma(1+x)+\gamma x}{x^{\frac{9}{4}}} dx = \frac{4\sqrt{2}\pi}{5}\zeta(\frac{5}{4})$ making the answer $\boxed{4+2+5+5+4=20}$

Integrating by parts (the asymptotic behaviour of $\ln\Gamma(x+1)$ near $x=0$ and as $x \to \infty$ makes this valid), we see that $\int_0^\infty \frac{\gamma x + \ln \Gamma(x+1)}{x^{\frac94}}\,dx \; = \; \tfrac45 \int_0^\infty \frac{\gamma + \psi(x+1)}{x^{\frac54}}\,dx$ But it is a standard integral (which can be proved by some contour integration) that $\int_0^\infty \frac{\gamma + \psi(x+1)}{x^\alpha}\,dx \; = \; -\pi \,\mathrm{cosec}\,(\alpha\pi)\zeta(\alpha) \hspace{2.5cm} 1 < \mathfrak{Re}\alpha < 2$ and hence the integral is equal to $\tfrac45 \times -\pi\,\mathrm{cosec}\big(\tfrac54\pi\big) \zeta\big(\tfrac54\big) \; = \; \frac{4\sqrt{2}\pi}{5}\zeta\big(\tfrac54\big)$ making the answer $4 + 2 + 5 + 5 + 4 = \boxed{20}$ .

Using IBP once the integral changes to $\displaystyle I=\frac 45\int_0^{\infty} \frac {\gamma +\psi(x+1)}{x^{\frac 54}}dx$ Using IBP once again this converts to $\displaystyle I=\frac {16}{5}\int_0^{\infty} \frac {\psi^{(1)}(x+1)}{x^{\frac 14}}dx$

Here $\psi(z)$ denotes Digamma function while $\psi^{(1)}(z)$ denotes Trigamma function.

Now using that $\displaystyle \psi^{(1)}(z+1)=\sum_{k\ge 0} \frac {(-z)^k}{k!} \zeta(k+2)\Gamma(k+2)$ (Where $\vert z\vert \lt 1$ ) Alongwith Ramanujan's Master Theorem we have $\displaystyle I=\frac {16}{5} \Gamma\left(\frac 34\right)\phi\left(-\frac 34\right)$ where $\displaystyle \phi(k)=\zeta(2+k)\Gamma(2+k)$

Using Euler's Reflection formula for Gamma function alongwith above result we get $\displaystyle I=\frac {4\sqrt 2 \pi}{5}\zeta\left(\frac 54\right)$