How should I integrate $ \int_0^1 \frac{(\log{x})^{2n-1}}{(1-x)\sqrt{x}} \, dx $? I'm interested in evaluating the integral ($n>0$)
$$ \int_0^1 \frac{(\log{x})^{2n-1}}{(1-x)\sqrt{x}} \, dx. $$
Unfortunately I have no idea how to tackle it. How should I try to evaluate it?
 A: $$
\begin{align}
\int_0^1\frac{\log(x)^{2n-1}}{(1-x)\sqrt{x}}\,\mathrm{d}x
&=4^n\int_0^1\frac{\log(x)^{2n-1}}{1-x^2}\,\mathrm{d}x\\
&=-4^n\int_0^\infty\frac{x^{2n-1}}{1-e^{-2x}}e^{-x}\,\mathrm{d}x\\
&=-4^n\int_0^\infty x^{2n-1}\left(e^{-x}+e^{-3x}+e^{-5x}+\dots\right)\,\mathrm{d}x\\
&=-4^n\Gamma(2n)\left(\frac1{1^{2n}}+\frac1{3^{2n}}+\frac1{5^{2n}}+\dots\right)\\
&=-4^n\Gamma(2n)\zeta(2n)\left(1-\frac1{4^n}\right)\\[6pt]
&=\left(1-4^n\right)\Gamma(2n)\zeta(2n)
\end{align}
$$
A: Calling $2n-1 = a$, and assuming $a > 0$, I would suggest to use the Geometric Series for the term
$$\frac{1}{1-x} \to \sum_{k = 0}^{+\infty} x^k$$
Since the integration is bounded in $[0, 1]$.
Hence:
$$\int_0^1 \sum_{k = 0}^{+\infty} x^k \frac{\log^a(x)}{\sqrt{x}}\ \text{d}x = \sum_{k = 0}^{+\infty} \int_0^1 x^{b} \log^a(x)\ \text{d}x$$
Where $b = k - \frac{1}{2}$
Now, with the help of a special function called Gamma Function, you can obtain
$$\int x^{b} \log^a(x)\ \text{d}x = \log ^{a+1}(x) (-(b+1) \log (x))^{-a-1} (-\Gamma (a+1,-(b+1) \log (x)))$$
And in your case
$$\int_0^1 x^{b} \log^a(x)\ \text{d}x = e^{i \pi  a} (b+1)^{-a-1} \Gamma (a+1)$$
As long as those conditions are satisfied:
$$\Re(b)>-1\land \Re(a)>-1$$
Then, you obtain a series (we left it apart to compute the integration):
$$\sum_{k = 0}^{+\infty} e^{i \pi  a} (b+1)^{-a-1} \Gamma (a+1)$$
Recalling that $b = k - \frac{1}{2}$ and $a = 2n-1$ we have:
$$\sum_{k = 0}^{+\infty} e^{i \pi  (2n-1)} \left(k + \frac{1}{2}\right)^{-2n} \Gamma (2n) = e^{i \pi  (2n-1)}\Gamma (2n)\sum_{k = 0}^{+\infty}\left(k + \frac{1}{2}\right)^{-2n}$$
The series
$$\sum_{k = 0}^{+\infty}\left(k + \frac{1}{2}\right)^{-2n} = \left(-1 + 2^{2n}\right)\zeta(2n)$$
Where $\zeta(\star)$ is the Riemann Zeta Function.
Finally, you obtain:
$$\boxed{\left(-1 + 2^{2n}\right)e^{i \pi  (2n-1)}\Gamma (2n)\zeta(2n)}$$
Last manipulation
Respecting the condition
$$\Re(x) > \frac{1}{2}$$
thanks to the exponential part we can also write the result in this form:
$$(-1)^{1 + 2n}(4^n - 1)\Gamma(2n)\zeta(2n)$$
And again:
$$\boxed{(1 - 4^n)\Gamma(2n)\zeta(2n)}$$
A: Let
$$ I(a)=\int_0^1\frac{x^a\log x}{(1-x)\sqrt{x}}dx. $$
Then
$$ I^{(2n-2)}(0)=\int_0^1\frac{(\log x)^{2n-1}}{(1-x)\sqrt{x}}dx. $$
Clear
\begin{eqnarray}
I(a)&=&\int_0^1\frac{x^a\log x}{(1-x)\sqrt{x}}dx\\
&=&\int_0^1\sum_{k=0}^\infty x^{k+a-\frac12}\log xdx\\
&=&\sum_{k=0}^\infty\int_0^1 x^{k+a-\frac12}\log xdx\\
&=&-\sum_{k=0}^\infty\frac{1}{(a+k+\frac12)^2}dx.
\end{eqnarray}
So
$$ I^{2n-2}(0)=-\sum_{k=0}^\infty\frac{(2n-1)!}{(k+\frac12)^{2n}}=(1-4^n)\Gamma(2n)\zeta（2n).$$
A: $$ I(n)=\int_0^1 \frac{(\log{x})^{2n-1}}{(1-x)\sqrt{x}} \, dx. $$
Change of variable : $x=e^{-t}$
$$ I(n)=\int_\infty^0 \frac{(-t)^{2n-1}}{(1-e^{-t})e^{-t/2}}\, (-e^{-t})dt = -\int_0^\infty \frac{ t^{2n-1} }{1-e^{-t}}\, e^{-t/2}dt $$
Laplace transform : $\Large \mathscr{L} \normalsize_t\left[\frac{ t^{2n-1} }{1-e^{-t}}\right](p) =  (2n-1)!\, \zeta(2n,p)$
$\zeta(2n,p)$ is the Hurwitz zeta function.
$$I(n)=-\Large \mathscr{L} \normalsize_t\left[\frac{ t^{2n-1} }{1-e^{-t}}\right]\left(\frac{1}{2}\right) = -(2n-1)!\, \zeta (2n,\frac{1}{2})$$
$$\int_0^1 \frac{(\log{x})^{2n-1}}{(1-x)\sqrt{x}} \, dx = -(2n-1)!\; \zeta (2n,\frac{1}{2})$$
$$\zeta (2n,\frac{1}{2})=(2^{2n}-1)\zeta(2n)$$
$\zeta(2n)$ is the Riemann zeta function.
