2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.