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

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