How do I evaluate the following definite integral: $\int^1_0\frac{x^2\ln x}{\sqrt{1-x^2}}dx$?

Question

$$\int^1_0\frac{x^2\ln x}{\sqrt{1-x^2}}dx$$ I tried using by parts but it didn't help. Also none of the properties of definite integral helps.

Answer 1

Let $x=\sin y$

$$I=\int_0^1\dfrac{x^2\ln x}{\sqrt{1-x^2}}dx=\int_0^{\pi/2}\sin^2y\ln(\sin y)dy$$

$$2I=\int_0^{\pi/2}\ln(\sin y)dy-\int_0^{\pi/2}\cos2y\ln(\sin y)dy$$

For the first half, see How do you evaluate the integral $\int_{0}^{\frac{\pi}{2}} \log(\sin x) dx$?

For the second,

$$\int\cos2y\ln(\sin y)dy=\ln(\sin y)\int\cos2y\ dy-\int\left(\dfrac{d\ \ln(\sin y)}{dy}\int\cos2y\ dy\right)dy=?$$

Hope you can take it from here!

Answer 2

Let $~I(k)=\displaystyle\int_0^1\frac{x^k}{\sqrt{1-x^2}}~dx.~$ Then our integral becomes $I'(2).~$ At the same time, a simple

trigonometric substitution yields $~I(k)=\displaystyle\int_0^\tfrac\pi2\sin^kx~dx.~$ But the latter is nothing more than

the famous Wallis integral, whose relation to the beta function is well-known. We do not need

to go through the former to show a link to the latter. Indeed, a simple substitution of the form

$t=x^2$ would have sufficed to establish such a connection. We have $I(k)=\dfrac\pi{k~B\bigg(\dfrac12~,~\dfrac k2\bigg)}$ .

Differentiating with regard to k, and using the relation between the beta and $\Gamma$ functions, we

will inevitably have to employ digamma functions, which are nothing more than harmonic

numbers in disguise. In particular, Euler's formula for extending the latter to non-natural

arguments will be of great help in evaluating $\psi_0\bigg(\dfrac32\bigg).~$ The final result should be

$$I'(2)~=~\dfrac\pi8\bigg(1-2\ln2\bigg).$$

Answer 3

Rewrite the integral as$$I(a)=\lim\limits_{a\to0}\frac {\partial}{\partial a}\int\limits_0^1dx\,\frac {x^{a+2}}{\sqrt{1-x^2}}$$And make the substitution $u=x^2$. You should get the resulting integral in terms of the beta function.

Answer 4

\begin{align}\int^1_0& \frac{x^2\ln x}{\sqrt{1-x^2}}dx \overset{x=\sin t }= \int_0^{\pi/2}\sin^2t\ln \sin t \ dt\\ =& \ -\int_0^{\pi/2}\int_0^1 \frac{y\cos^2 t}{1+y^2\cot^2t}dy\ dt\\ =& -\frac\pi4 \int_0^1 \frac y{(1+y)^2}dy =\frac\pi8 -\frac\pi4 \ln2 \end{align}

Answer 5

$$\begin{align*} & \int_0^1 \frac{x^2\ln(x)}{\sqrt{1-x^2}} \, dx \\ &= - \int_0^1 \left[\int_x^1 \frac{dy}y\right] \frac{x^2}{\sqrt{1-x^2}} \, dx \tag1 \\ &= - \int_0^1 \left[\int_0^y \frac{x^2}{\sqrt{1-x^2}} \, dx\right] \frac{dy}y \tag2 \\ &= \frac12 \int_0^1 \sqrt{1-y^2} \, dy + \int_0^1 \left(\arctan\left(\frac{1+\sqrt{1-y^2}}y\right) - \frac\pi2\right)\,\frac{dy}y \\ &= \frac\pi8 + \int_1^\infty \left(\frac\pi2-\arctan(z)\right)\,\frac{1+z^2}{2z} \cdot 2\frac{1-z^2}{(1+z^2)^2} \, dz \tag3 \\ &= \frac\pi8 + \int_1^\infty \arctan\left(\frac1z\right) \cdot \frac{1-z^2}{z(1+z^2)} \, dz \tag4 \\ &= \frac\pi8 - \int_0^1 \arctan(w) \cdot \frac{1-w^2}{w(1+w^2)} \, dw \tag5 \\ &= \frac\pi8 + \frac\pi4\ln(2) + \int_0^1 \frac{\ln(w) - \ln(1+w^2)}{1+w^2} \, dw \\ &= \boxed{\frac\pi8 - \frac\pi4\ln(2)} \end{align*}$$

---

• $(1)$ : integral definition of $\ln(x)$
• $(2)$ : apply Fubini's theorem
• $(3)$ : substitute $z=\dfrac{1+\sqrt{1-y^2}}y \iff y=\dfrac{2z}{1+z^2}$
• $(4)$ : recall $\arctan(z)+\arctan\left(\dfrac1z\right)=\dfrac\pi2$ for $z>0$
• $(5)$ : substitute $w=\dfrac1z$
• $(6)$ : integrate by parts

Evaluation of the remaining integrals can be done using the results from here and here.

Answer 6

Integration by parts: IBP. $\begin{align}\require{enclose} \int_0^1\frac{x^2\ln x}{\sqrt{1-x^2}}dx&\stackrel{IBP}=\enclose{updiagonalstrike}{\left.\left(\frac12\arcsin x-\frac12x\sqrt{1-x^2}\right)\ln x\right\rvert_0^1}-\int_0^1\left(\frac12\frac{\arcsin x}x-\frac12\sqrt{1-x^2}\right)dx\\ &=\frac\pi 8-\frac12\int_0^1\frac{\arcsin x}xdx\\ &\stackrel{IBP}=\frac\pi 8+\frac12\int_0^1\frac{\ln x}{\sqrt{1-x^2}}dx\\ &\stackrel{x\rightarrow\sin x}=\frac\pi 8+\frac12\int_0^{\pi/2}\ln\sin x\,dx\\ &=\frac\pi 8-\frac\pi 4\ln2. \end{align}$