Calculation of $k+l$ 
If $\displaystyle \int^{1}_{0}\frac{\ln^{1729}(x)}{(x-1)\sqrt{x}}dx=k \int^{\infty}_{0}\frac{x^l}{e^x-1}dx$. then $\displaystyle k+l$ is

$\bf{Attempt:}$ Substitute $x=\sin^2 \theta$ in $\displaystyle \int^{1}_{0}\frac{\ln^{1729}(x)}{(x-1)\sqrt{x}}dx$ 
$$=-\int^{\frac{\pi}{2}}_{0}\frac{\ln^{1729}(\sin^2 \theta)}{\cos^2 \theta \sin \theta}\cdot 2\sin \theta \cos \theta d\theta=-2\int^{\frac{\pi}{2}}_{0}\frac{\ln^{1729}(\sin^2 \theta)}{\cos \theta}d\theta$$
could some help me how to solve it, thanks
 A: Let $x = e^y$ then
$$
\displaystyle \int^{1}_{0}\frac{\ln^{1729}(x)}{(x-1)\sqrt{x}}dx = \displaystyle \int^{0}_{-\infty}\frac{y^{1729} e^{y/2}}{(e^y-1)}dy
$$
Now resubstitute $y = -2 x$ to get 
$$
2^{1730} \displaystyle \int_{0}^{\infty}\frac{x^{1729} e^{-x}}{(1-e^{-2x})}dx =2^{1730} \displaystyle \int_{0}^{\infty}\frac{x^{1729} e^{x}}{e^{2x}-1}dx
$$
Note 
$$
\frac{e^x}{e^{2x}-1} = \frac{1}{e^x-1} - \frac{1}{e^{2x} -1}
$$
So we get
$$
2^{1730} \displaystyle \int_{0}^{\infty}\frac{x^{1729} }{e^x-1}dx - 2^{1730} \displaystyle \int_{0}^{\infty}\frac{x^{1729}}{e^{2x} -1}dx
$$
In the last term, replace $y = 2x$ which gives
$$
2^{1730} \displaystyle \int_{0}^{\infty}\frac{x^{1729} }{e^x-1}dx -  \displaystyle \int_{0}^{\infty}\frac{y^{1729}}{e^{y} -1}dx = (2^{1730}  - 1) \displaystyle \int_{0}^{\infty}\frac{x^{1729} }{e^x-1}dx 
$$
So we have $k = 2^{1730}  - 1$ and $l = 1729 $, giving $k+l = 2^{1730}  + 1728$.
A: Since for any $\alpha >-1$ and any $\beta\in\mathbb{N}$ we have $$\int_{0}^{1}x^{\alpha}\log(x)^\beta\,dx=\left.\frac{d^\beta}{dz^\beta}\int_{0}^{1}x^{\alpha+z}\,dx\, \right|_{z=0^+}=\frac{(-1)^\beta\cdot\beta!}{(\alpha+1)^{\beta+1}}$$
by expanding $\frac{1}{(x-1)\sqrt{x}}$ as $-\left(\frac{1}{\sqrt{x}}+\sqrt{x}+x\sqrt{x}+\ldots\right)$ we get
$$ \int_{0}^{1}\frac{\log^{1729}(x)}{(x-1)\sqrt{x}}\,dx=1729!\sum_{n\geq 0}\frac{1}{\left(\frac{2n+1}{2}\right)^{1730}}=1729!\,(2^{1730}-1)\,\zeta(1730).$$
On the other hand 
$$\int_{0}^{+\infty}\frac{x^l}{e^x-1}\,dx=l!\,\zeta(l+1)$$
hence we have to pick $l=1729$ and $k=2^{1730}-1$.
