Integral $\int_0^1 \frac{\sqrt x \ln x} {x^2 - x+1}dx$

Question

I am trying to evaluate $$I=\int_0^1 \frac{\sqrt x \ln x} {x^2 - x+1}dx=\int_0^1 \frac{\sqrt x (1+x)\ln x} {1+x^3}dx$$ Now if we expand into geometric series: $$I=\sum_{n=0}^{\infty} (-1)^n \int_0^1 (x^{3/2}+x^{1 /2})x^{3n}\ln x dx$$ Also since $$I(k) =\int_0^1 x^kdx=\frac{1} {k+1}$$ Giving: $$I'(k) =\int_0^1 x^k\ln x dx=-\frac{1} {(k+1)^2 }$$ so using this we get $$I=\sum_{n=0}^{\infty} (-1)^{n+1}\left(\frac{1} {(6n+3)^2 }+\frac{1} {(6n+1) ^2 }\right)$$ Now when I plug this into wolfram-alpha the result differs from the value of the integral, also if I multiply by a half it is really close to it. Where did I go wrong?

Edit: Looks like I forgot a 2 in the denominator and to add $+1$ from $I'(k) $ and the correct series should be:$$I=\frac{4}{36}\sum_{n=0}^{\infty} (-1)^{n+1}\left(\frac{1} {(n+5/6)^2 }+\frac{1} {(n+1/2) ^2 }\right)$$ The second one is just $-4G$ where $G$ is the Catalan constant and can you show me how to transform the sum into a closed form? Trigamma or hurwitz zeta function as wolfram alpha gives as a solution. Many thanks in advance!

Answer 1

Multiply the numerator and denominator by $x+1$ and expand the linear term to get

$$\begin{align*}I & =\int\limits_0^1dx\,\frac {x^{1/2}\log x}{1+x^3}+\int\limits_0^1dx\,\frac {x^{3/2}\log x}{1+x^3}\\ & =-\sum\limits_{n\geq0}(-1)^{n}\int\limits_0^1dx\, x^{3n+1/2}\log x-\sum\limits_{n\geq0}(-1)^{n}\int\limits_0^1dx\, x^{3n+3/2}\log x\end{align*}$$

Integration on both integrals gives us

$$I=\sum\limits_{n\geq0}\frac {(-1)^n}{(3n+3/2)^2}+\sum\limits_{n\geq0}\frac {(-1)^n}{(3n+5/2)^2}$$

The first integral can be evaluated by factoring out a three from the denominator. The sum then becomes the infinite sum for $G$, Catalan's constant

$$\begin{align*}\sum\limits_{n\geq0}\frac {(-1)^n}{(3n+3/2)^2} & =\frac 49\sum\limits_{n\geq0}\frac {(-1)^n}{(2n+1)^2}\\ & =\frac 49G\end{align*}$$

The second sum can be rewritten in terms of the Hurwitz Zeta function, and consequently, the polygamma function. First, we factor out a three from the denominator like before, and expanding the series and adding and subtracting the two sequences together, we get

$$\begin{align*}\sum\limits_{n\geq0}\frac {(-1)^n}{(3n+5/2)^2} & =\frac 19\sum\limits_{n\geq0}\frac {(-1)^n}{(n+5/6)^2}\\ & =\frac 1{36}\sum\limits_{n\geq0}\frac 1{(n+5/12)^2}-\frac 1{36}\sum\limits_{n\geq0}\frac 1{(n+11/12)^2}\\ & =\frac 1{36}\left[\zeta\left(2,\frac 5{12}\right)-\zeta\left(2,\frac {11}{12}\right)\right]\end{align*}$$

Note that in general

$$\sum\limits_{n\geq0}\frac {(-1)^n}{(n+a)^2}=\frac 14\left[\zeta\left(2,\frac a2\right)-\zeta\left(2,\frac {a+1}2\right)\right]$$

Putting everything together, and using the fact that $\zeta(2,a)=\psi'(a)$, then the integral evaluates as

$$\int\limits_0^1dx\,\frac {\sqrt{x}\log x}{x^2-x+1}\color{blue}{=\frac 1{36}\psi'\left(\frac {11}{12}\right)-\frac 1{36}\psi'\left(\frac 5{12}\right)-\frac {4G}9}$$

Answer 2

$$ \frac{\sqrt{x}}{x^2-x+1} = \sum_{k=0}^\infty (-1)^k \left(x^{3k+1/2} + x^{3k + 3/2}\right) $$ so $$ \frac{\sqrt{x} \ln(x)}{x^2-x+1} = \left.\sum_{k=0}^\infty (-1)^k \dfrac{d}{dp}\left(x^{3k+1/2+p} + x^{3k+3/2+p}\right)\right|_{p=0}$$

$$ \eqalign{\int_0^1 \frac{\sqrt{x} \ln(x)}{x^2-x+1} \; dx &= \sum_{k=0}^\infty (-1)^k \left.\dfrac{d}{dp} \left( \frac{1}{3k+3/2+p} + \frac{1}{3k+5/2+p} \right)\right|_{p=0}\cr &= \sum_{k=0}^\infty (-1)^{k+1} \left( \frac{1}{(3k+3/2)^2} + \frac{1}{(3k+5/2)^2}\right)\cr}$$

EDIT: In terms of $\zeta(2,v) = \sum_{j=0}^\infty 1/(j+v)^2$, we can write this as $$ \frac{1}{36} \left(-\zeta(2,1/4) -\zeta(2,5/12) + \zeta(2,3/4)+\zeta(2,11/12)\right) $$ since for $k=2j$, $3k+3/2 = 6 (j+1/4)$, for $k=2j+1$, $3k+3/2 = 6 (j+3/4)$, etc.