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

  • $\begingroup$ Come on, that's hardly an honest attempt. $\endgroup$ – uniquesolution Feb 25 '19 at 14:32
  • 1
    $\begingroup$ 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 $x=1$ $\endgroup$ – 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.

  • $\begingroup$ Listen to Kluas! $\endgroup$ – user150203 Feb 26 '19 at 6:16

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