How to Solve Complex Logarithmic Integrals?

[sutupidmath]

The coefficient of your first term on RHS should be 4 not 2.



You have a sign error; it should be +2x not -2x.




Instead of using the series (which doesn't converge over the interval we are interested in, it is better to use the definition of PolyLogarithm Mute gave to show that

\int \frac{\ln(x+1)}{x}dx=-\operatorname{Li}_2(-x)

And hence:

\int \ln(x)\ln(x+1)dx=\ln(x)[(x+1)\ln(x+1)-x]+2x-(x+1)\ln(x+1)+\operatorname{Li}_2(-x)



Again; avoid the divergent series by showing that

\int \frac{\ln(x+2)}{x}dx=\ln(2)\ln(x)-\operatorname{Li}_2(\frac{-x}{2})

And Hence:

\int \ln(x)\ln(x+2)dx=\ln(x)[(x+2)\ln(x+2)-x]-(x+2)\ln(x+2)+2x-\ln(2)\ln(x)+\operatorname{Li}_2\left( \frac{-x}{2}\right)



Good.



Again, avoid the series by showing

\int \frac{\ln(x+2)}{x+1}dx=-\operatorname{Li}_2(-(x+1))

And Hence:

\int \ln(x+1)\ln(x+2)dx=\ln(x+1)[(x+2)\ln(x+2)-(x+1)]-(x+2)\ln(x+2)+2x+\operatorname{Li}_2(-(x+1))



There is a small error, you should get:

\int \ln^2(x+2)dx=(x+2)\ln^2(x+2)-2(x+2)\ln(x+2)+2x

Fix those errors and then add up your terms and cancel as much as possible to show that

\int \ln^2\left(\frac{x^2}{x^2+3x+2}\right) dx=8\ln(2)\ln(x) + 4 x\ln^2(x)- 4 (x+1)\ln(x)\ln(x+1) + (x+1)\ln^2(x+1) - 4 (x+2)\ln(x)\ln(x+2)

+2 (x+2)\ln(x+1)\ln(x+2)+ (x+2)\ln^2(x+2) + 2\operatorname{Li}_2(-(x+1))-4\operatorname{Li}_2(-x)-8\operatorname{Li}_2\left(-\frac {x} {2} \right)

After you do that, I'll help you through incorporating the limits.

I have few comments:

1. Shouldn't we on the last result have 4x^2ln^2(x) instead of plain x, without square?

2. also, here

\int \frac{ln(x+2)}{x}dx=\int \frac{ln[2(1-(-\frac{x}{2}))]}{x}dx=\int \frac{ln(2)}{x}+\int \frac{[1-(-\frac{x}{2})]}{-2*\frac{-x}{2}}=ln(2)ln(x)+\frac{1}{2}Li_2(\frac{-x}{2})

If so then we would have a slight change on our final result: i.e.

we would have 4ln(2)ln(x)+2Li(-\frac{x}{2})

Or, am i missing something here?


Other than these, i managed to bring the expression to match that of yours.

 

[gabbagabbahey]

I have few comments:

1. Shouldn't we on the last result have 4x^2ln^2(x) instead of plain x, without square?

Nope, I overlooked an error in your first integral; you should have

\int \ln^2(x)dx=x\ln^2(x)-2x\ln(x)+2x

That gets rid of the x squared thing

2. also, here

\int \frac{ln(x+2)}{x}dx=\int \frac{ln[2(1-(-\frac{x}{2}))]}{x}dx=\int \frac{ln(2)}{x}+\int \frac{[1-(-\frac{x}{2})]}{-2*\frac{-x}{2}}=ln(2)ln(x)+\frac{1}{2}Li_2(\frac{-x}{2})

Nope, you need to use

\operatorname{Li}_2\left( \frac{-x}{2}\right)=-\int \frac{\ln[1-(-\frac{x}{2})]}{\frac{-x}{2}} d\left( \frac{-x}{2} \right)\neq -\int \frac{\ln[1-(-\frac{x}{2})]}{\frac{-x}{2}} dx