# If $\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x)^2}dx+k\int^{1}_{0}\frac{\ln(1-x)}{x}dx=0.$ Find $k$.

If $$\displaystyle \int^{\infty}_{0}\frac{\ln^2(x)}{(1-x)^2}dx+k\int^{1}_{0}\frac{\ln(1-x)}{x}dx=0.$$ Find $$k$$.

Try: Let $$\displaystyle I =\int^{\infty}_{0}\ln^2(x)\cdot \frac{1}{(1-x)^2}dx$$

Integrate by parts

So $$I =-\ln^2(x)\cdot \frac{1}{1-x}\bigg|^{\infty}_{0}+2\int^{\infty}_{0}\frac{\ln(x)}{1-x}dx$$

Now Could not go ahead, Could some help me to solve it, Thanks

• Come on, that's hardly an honest attempt. – uniquesolution Feb 25 '19 at 14:32
• Note that $\int_{0}^{\infty}(\frac{ln^2(x)}{(1-x)^2}) = \int_{0}^{1}(\frac{ln^2(x)}{(1-x)^2}) + \int_{1}^{\infty}(\frac{ln^2(x)}{(1-x)^2})$ because the integrand is not defined if $x=1$ – LuxGiammi Feb 25 '19 at 14:36

Not sure if this is the fastest way, but the substitution $$y = \frac{1}{x}$$ followed by partial integration yields $$\int_0^1 \frac{\ln(x)}{1-x} \, \mathrm{d}x = \int_1^{\infty} \frac{\ln(y)}{1-y} \frac{1}{y} \, \mathrm{d}y = -\int_1^{\infty} \frac{\ln(y)}{y(1-y)} + \frac{\ln(y)^2}{(1-y)^2} \, \mathrm{d}y,$$ hence $$2\int_0^1 \frac{\ln(x)}{1-x} \, \mathrm{d}x = -\int_1^{\infty} \frac{\ln(y)^2}{(1-y)^2} \, \mathrm{d}y.$$ Now observe that by the same substitution $$\int_1^{\infty} \frac{\ln(y)^2}{(1-y)^2} \, \mathrm{d}y = \int_0^1 \frac{\ln(y)^2}{(1-y)^2} \, \mathrm{d}y,$$ hence $$4\int_0^1 \frac{\ln(x)}{1-x} \, \mathrm{d}x = -\int_0^{\infty} \frac{\ln(y)^2}{(1-y)^2} \, \mathrm{d}y,$$ so the answer should be $$k = 4$$. I left some details for you to fill in.