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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

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    $\begingroup$ What exactly is the problem? $\endgroup$ – tired Nov 23 '15 at 17:26
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ A very similar question. $\endgroup$ – Lucian Nov 23 '15 at 17:51
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An alternative:

Consider

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

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

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

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