Evaluation of $\displaystyle \int_{0}^{1}\frac{x\ln (x)}{\sqrt{1-x^2}}dx$

$\bf{My\; Try::}$ Let $\displaystyle I = \int_{0}^{1}\frac{x\ln x}{\sqrt{1-x^2}}dx\;,$ Put $x=\cos \phi\;,$ Then $dx = -\sin \phi d\phi$

and Changing Limit, We get

$$\displaystyle I = -\int_{\frac{\pi}{2}}^{0}\cos \phi \cdot \ln(\cos \phi )d\phi = \int_{0}^{\frac{\pi}{2}} \ln(\cos \phi)\cdot \cos \phi d\phi$$

Now Using Integration by parts, We get

$$\displaystyle I = \left[\ln(\cos \phi)\cdot \sin \phi\right]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}\frac{\sin^2 \phi}{\cos \phi}d\phi$$

So $$\displaystyle I = \left[\ln(\cos \phi)\cdot \sin \phi\right]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}\frac{(1-\cos^2 \phi)}{\cos \phi}d\phi$$

So $$\displaystyle I = \left[\ln(\cos \phi)\cdot \sin \phi\right]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}\sec \phi d\phi-\int_{0}^{\frac{\pi}{2}}\cos \phi d\phi$$

So $$\displaystyle I = \left[\ln(\cos \phi)\cdot \sin \phi\right]_{0}^{\frac{\pi}{2}}+\left[\ln\left|\sec \phi+\tan \phi\right|\right]_{0}^{\frac{\pi}{2}}-\left[\sin \phi\right]_{0}^{\frac{\pi}{2}}$$

So $$\displaystyle I = \left[\ln(\cos \phi)\cdot \sin \phi\right]_{0}^{\frac{\pi}{2}}+\left[\ln\left|\sec \phi+\tan \phi\right|\right]_{0}^{\frac{\pi}{2}}-1$$

Now How can I solve after that, Help Required, Thanks

  • 1
    $\begingroup$ What exactly is the problem? $\endgroup$ – tired Nov 23 '15 at 17:26
  • 2
    $\begingroup$ It looks like you already solved it? $\endgroup$ – Mankind Nov 23 '15 at 17:30
  • $\begingroup$ actually here How can I put that upper and lower limit. means in first part, $\ln(0)\cdot 1 = -\infty$ and in second part $\ln|\frac{2}{0}|$ $\endgroup$ – juantheron Nov 23 '15 at 17:31
  • 4
    $\begingroup$ I think that you have a problem in the formula just after " integrating by parts We get": The integrated part does not exist (infinite limit if $\phi \to \pi/2$ ) and the integral neither (problem at $\pi/2$). To correct choose $\sin(\phi)-1$ as a primitive for $\cos(\phi)$. Then the integrated part is $0$, and your integral is convergent. $\endgroup$ – Kelenner Nov 23 '15 at 17:33
  • 1
    $\begingroup$ A very similar question. $\endgroup$ – Lucian Nov 23 '15 at 17:51

An alternative:


$$ J(a)=\int_0^{\pi/2} \cos^a(\phi)d\phi $$

differentiating w.r.t $a$ gives us $$ \partial_a J(a)\big|_{a=1}=-I $$

But on the other hand $J(a)$ is just a Wallis integral and therefore

$$ I=-\frac{\sqrt{\pi }}{2}\partial_a\left( \frac{\Gamma \left(\frac{a+1}{2}\right)}{\Gamma \left(\frac{a}{2}+1\right)}\right)\big|_{a=1} $$

which yields the desiered result after using a special value of the Digamma function

$$ I=\log(2)-1 $$


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.