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I am trying to calculate:

$$\int_0^1 \frac{\ln(1-x+x^2)}{x-x^2}dx$$

I am not looking for an answer but simply a nudge in the right direction. A strategy, just something that would get me started.

So, after doing the Taylor Expansion on the $\ln(1-x+x^2)$ ig to the following: Let $x=x-x^2$ then $\ln(1-x)$ then, \begin{align*} =&-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-...\\ =&-(x-x^2)-\frac{(x-x^2)^2}{2}-...\\ =&-x(1-x)+\frac{x^2}{2}(1-x)^2-\frac{x^3}{3}(1-x)^3\\ \text{thus the pattern is:}\\ =&\frac{x^n(1-x)^n}{n} \end{align*} Am I right?

Then our Integral would be: $$\sum_{n=0}^{\infty} \frac{1}{n+1} \int_0^1 x^n(1-x)^n$$

Am I on the right track? Suggestions, tips, comments?

$\underline{NEW EDIT:}$

SO after integrating the function I got the following after a couple of iterations: \begin{align*} \frac{n(n-1)...1}{(n+1)(n+2)...(2n)}\int_0^1 x^{2n} dx \end{align*} This shows a pattern: \begin{align*} =&\frac{(n!)^2}{(2n)!} (\frac{1}{2n+1})\\ =& \frac{(n!)^2}{(2n+1)!} \end{align*} So my question is, what to do from here. I have done all this but still have no clue how to actually solve the integral. Can somebody shed some light on this! Thanks

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4  
Expand the log into a Taylor series about $x-x^2$, perform the integration, and then recognize the resulting series as an expansion of a (sort of) well-known function evaluated at a particular value. –  Ron Gordon Apr 16 at 16:59
    
@Ron Gordon Mmmmh. Very interesting. I will work on it and will post what I come up with. Thanks –  user2233524 Apr 16 at 20:28
    
Ok guys I added some stuff. Still need some help. Thanks –  user2233524 Apr 18 at 19:30
    
@user2233524 $\int_0^1 \frac{\ln(1-x+x^2)}{x-x^2}dx$ is less than zero, whereas $\sum_{n=0}^{\infty} \frac{1}{n+1} \int_0^1 x^n(1-x)^n$ is clearly positive, so you must have made a mistake somewhere. They seem to have equal magnitude though, so it's probably just a simple sign error. –  David H Apr 18 at 21:47

2 Answers 2

Just to simplify the things, make the change of variables $s=2x-1$. The integral then reduces to $$I=2\int_{-1}^1\frac{\ln\frac{3+s^2}{4}}{1-s^2}ds.\tag{1}$$ The antiderivative of any expression of the type $\displaystyle\frac{\ln P(x)}{Q(x)}$ is computable in terms of dilogarithms, essentially due to $$\displaystyle \int\frac{\ln(a-x)}{x+b}dx=\mathrm{Li}_2\left(\frac{a-x}{a+b}\right)+\ln(a-x)\ln\frac{x+b}{a+b}.\tag{2}$$ Hence the answer can be certainly expressed in terms of dilogarithm values.

Let us spell this out more explicitly. It is convenient to integrate once by parts and rewrite (1) as \begin{align}I=&-\int_{-1}^1\frac{2s}{3+s^2}\ln\frac{1+s}{1-s}ds= 4\Re\int_{-1}^1\frac{\ln(1-s)}{s+i\sqrt3}ds. \end{align} Applying (2), this reduces to $$I=-4\Re\,\mathrm{Li}_2\left(e^{i\pi/3}\right)=-\frac{\pi^2}{9},$$ where at the last step we have used that for $z\in(0,1)$ one has $$\Re\,\mathrm{Li}_2\left(e^{2i\pi z}\right)=\pi^2\left(z^2-z+\frac16\right).$$

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I don't understand the substitution you made in the integral. –  user2233524 Apr 19 at 20:25
    
@user2233524 As $x^2-x= \left(x-\frac12\right)^2-\frac14$, I set $x-\frac12=\frac{s}{2}$ $\Longleftrightarrow$ $s=2x-1$. –  O.L. Apr 19 at 20:29
    
It is still unclear to me how you get the equalities after "It is convenient to integrate once by parts and rewrite (1) as." Where does the imaginary part come form? –  user103828 Apr 20 at 8:21
1  
@user103828 The first equality comes from $\left(\ln\frac{1+s}{1-s}\right)'=\frac{2}{1-s^2}$ and integration by parts. The second equality comes from the fact that $\left(\ln(3+s^2)\right)'=\frac{2s}{3+s^2}$ is odd function (so that we can replace $\ln\frac{1+s}{1-s}$ by $-2\ln(1-s)$) and the decomposition $\frac{2s}{3+s^2}=\frac{1}{s+i\sqrt3} + \frac{1}{s-i\sqrt3} = 2\Re \frac{1}{s+i\sqrt3}$. –  O.L. Apr 20 at 8:40
    
@O.L. +1, do you have proof/reference for (2). Seems like a useful thing to know. thanks –  Integrals Apr 21 at 16:19

Differentiation under the integral sign can be applied to this.

Consider

\begin{align} I(a)&=\int_0^1 \, \frac{\ln{(1-a\,(x-x^2))}}{x-x^2}\, dx \tag 1\\ \frac{\partial}{\partial a} I(a)&=\int_0^1\, -\frac{1}{1-a\,(x-x^2)}\, dx \\ &= -\frac{4 \, \sqrt{-y^{2} + 4 \, y} \arctan\left(\frac{\sqrt{-y^{2} + 4 \, y}}{y - 4}\right)}{y^{2} - 4 \, y}\tag 2 \end{align}

Integrating $(2)$ gives us (I used Sage for that)

\begin{align} I(a)=\sqrt{a} \sqrt{-a + 4} \arcsin\left(\frac{1}{2} \, a - 1\right) + 2 \, \arcsin\left(\frac{1}{2} \, a - 1\right)^{2} - 4 \, \arctan\left(\frac{{\left(a - 2\right)} \sqrt{a} \sqrt{-a + 4}}{a^{2} - 4 \, a}\right) \arctan\left(\frac{\sqrt{-a + 4}}{\sqrt{a}}\right) + 4 \, \arctan\left(\frac{\sqrt{-a + 4}}{\sqrt{a}}\right)^{2} + 4 \, \arcsin\left(\frac{1}{2} \, a - 1\right) \arctan\left(\frac{\sqrt{-a^{2} + 4 \, a}}{a - 4}\right) - \sqrt{-a^{2} + 4 \, a} \arcsin\left(\frac{1}{2} \, a - 1\right) + C \\ \tag 3 \end{align}

Setting $a=0$ in $(2)$ and $(3)$, $C=-\frac{\pi^2}{2}$, hence:

\begin{align} I(a)=\sqrt{a} \sqrt{-a + 4} \arcsin\left(\frac{1}{2} \, a - 1\right) + 2 \, \arcsin\left(\frac{1}{2} \, a - 1\right)^{2} - 4 \, \arctan\left(\frac{{\left(a - 2\right)} \sqrt{a} \sqrt{-a + 4}}{a^{2} - 4 \, a}\right) \arctan\left(\frac{\sqrt{-a + 4}}{\sqrt{a}}\right) + 4 \, \arctan\left(\frac{\sqrt{-a + 4}}{\sqrt{a}}\right)^{2} + 4 \, \arcsin\left(\frac{1}{2} \, a - 1\right) \arctan\left(\frac{\sqrt{-a^{2} + 4 \, a}}{a - 4}\right) - \sqrt{-a^{2} + 4 \, a} \arcsin\left(\frac{1}{2} \, a - 1\right)-\frac{\pi^2}{2} \end{align}

Therefore, the required answer is $$ I(1)=-\frac{\pi^2}{9}\approx -1.09662271123215 $$

=== Update ===

There is a simpler general form as I suspected:

Instead of $(1)$, consider $I(a)$ to be

\begin{align} I(a)&=\int_0^1 \, \frac{\ln{(1+a\,(x-x^2))}}{x-x^2}\, dx \end{align}

Applying the differentiation under the integral sign now yields a nice simpler form:

\begin{align} I(a)&=\ln\left( \frac{a+2 -\sqrt{a^{2} + 4 \, a}}{2}\right)^{2} \end{align}

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Do you have an analytical proof of (2)? A numerical answer isn't much of a proof. The method of differentiation under the integral sign is very nice here though, +1 –  Integrals Apr 21 at 16:16
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I do not know how $(2)$ is evaluated by hand, there may be an alternative simpler answer since sometimes CAS evaluates to more complicated expressions. The differentiation under the integral sign makes possible to evaluate some integrals which otherwise cannot be evaluated even by a CAS. –  gar Apr 21 at 17:18

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