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

Question

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

Answer 1

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

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

Answer 2

Substitute $t^2=1-x^2$

\begin{align} \int_{0}^{1}\frac{x\ln (x)}{\sqrt{1-x^2}}dx &=\frac12 \int_0^1 \ln(1-t^2)d(t-1)\\ &\overset{ibp} =-\int_0^1\frac t{1+t}dt =\ln2-1 \end{align}

Answer 3

$$\begin{align*} & \int_0^1 \frac{x \ln (x)}{\sqrt{1-x^2}} \, dx \\ &= \int_0^1 \left[- \int_x^1 \frac{dy}{y} \right] \frac x{\sqrt{1-x^2}} \, dx \\ &= - \int_0^1 \left[\int_0^y \frac x{\sqrt{1-x^2}} \, dx\right] \, \frac{dy}y \\ &= \int_0^1 \frac{\sqrt{1-y^2} - 1}y \, dy \\ &= \ln(2)-1 \end{align*}$$

Answer 4

Since $\displaystyle\int_0^{\pi/2}\cos\phi\ln(\cos\phi)\mathrm d\phi$ is an improper integral due to the upper bound $\frac\pi2$, at the end we can't simply plug in $\phi=\frac\pi2$ in the anti-derivative; rather we must take the limit $\phi\to\frac\pi2$. So, the given integral is

\begin{align}I&=\lim_{\phi\to\frac\pi2}\sin\phi\ln(\cos\phi)+\ln(\sec\phi+\tan\phi)-1\\&=\lim_{x\to0}\sqrt{1-x^2}\ln x+\ln\left(1+\sqrt{1-x^2}\right)-\ln x-1\\&=\ln2-1+\lim_{x\to0}\left(-\frac{x^2}2+O(x^4)\right)\ln x\\&=\ln2-1\end{align}

Answer 5

Disclaimer: This is like Quanto's answer but breaks things down more

Let $u=\sqrt{1-x^2}$. Then $du=-\frac{xdx}{\sqrt{1-x^2}}$. Thus we have that $$I=\int_0^1\ln(\sqrt{1-u^2})du=\frac12\left(\int_0^1\ln(1-u)+\ln(1+u)du\right)$$The first integral is $-1$ and the second one is $2\ln2-1$. Thus $I=\ln2-1$.

Answer 6

Although too long, a solution by using series can be done. Let's make some preliminaries before the solution: $$\frac1{\sqrt{1-t}}=\sum_{n=0}^\infty c_nt^n,\hspace{1cm}\text{for |t|<1 where $c_n={2n\choose n}2^{-2n}$}$$ By taking integral, adjusting the indefinite integral constant and dividing by $t$, we have $$\frac{2-2\sqrt{1-t}}t=\sum_{n=0}^\infty \frac{c_n}{n+1}t^n$$ By taking integral, adjusting the indefinite integral constant again, we have $$4\ln(1+\sqrt{1-t})-4\sqrt{1-t}+4-4\ln2=\sum_{n=0}^\infty \frac{c_n}{(n+1)^2}t^{n+1}$$ from which we get $$\sum_{n=0}^\infty \frac{c_n}{(n+1)^2}=4-4\ln2$$ by letting $t=1$.

The solution: $$\begin{align} \int_0^1\frac{x\ln x}{\sqrt{1-x^2}}dx&=\int_0^1x\ln x\sum_{n=0}^\infty c_nx^{2n}\\ &=\sum_{n=0}^\infty c_n\int_0^1 x^{2n+1}\ln x dx\\ &=\sum_{n=0}^\infty c_n\left(-\frac1{(2n+2)^2}\right)\\ &=-\frac14\sum_{n=0}^\infty\frac{c_n}{(n+1)^2}\\ &=-\frac14(4-4\ln2)=\ln2-1 \end{align}$$ where we used the fact that $\int_0^1x^m\ln xdx=-\frac1{(m+1)^2}$, which is obtained by integration by parts.