1

$$\int_{0}^{1}\dfrac{x^2\ln x}{\sqrt{1-x^2}}dx$$

My attempt is as follows:

$$x=\sin\theta\Rightarrow dx=\cos\theta d\theta$$

$$\int_{0}^{\frac{\pi}{2}}\sin^2\theta\ln (\sin\theta)d\theta=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}(1-\cos2\theta)\ln(\sin\theta)d\theta$$ $$=\frac{1}{2}\left(\int_{0}^{\frac{\pi}{2}}\ln(\sin\theta)d\theta-\int_{0}^{\frac{\pi}{2}}\cos(2\theta)\ln(\sin\theta)d\theta\right)=\frac{1}{2}\left(\dfrac{-\pi\ln(2)}{2}-I'\right)$$

$$I'=\int_{0}^{\frac{\pi}{2}}\cos(2\theta)\ln(\sin\theta)d\theta=\int_{0}^{\frac{\pi}{2}}\cos\left(\pi-2\theta\right)\ln(\cos\theta)d\theta$$

$$\Rightarrow 2I'=\int_{0}^{\frac{\pi}{2}}\cos(2\theta)\ln(\tan\theta)d\theta$$

As $\cos(2\theta)\ln(\tan\theta)d\theta=\cos\left(2\left(\frac{\pi}{2}-\theta\right)\right)\ln\left(\tan\left(\dfrac{\pi}{2}-\theta\right)\right)d\theta$

$$I'=\int_{0}^{\frac{\pi}{4}}\cos(2\theta)\ln(\tan\theta)d\theta$$

$$I'=\int_{0}^{\frac{\pi}{4}}\frac{1-\tan^2\theta}{1+\tan^2\theta}\ln(\tan\theta)d\theta$$

$$\tan\theta=t$$ $$I'=\int_{0}^{1}(1-t^2)\ln(t)dt$$

Applying integration by parts

$$I'=-\lim_{t\to0}\ln(t)\cdot\left(t-\frac{t^3}{3}\right)-\int_{0}^{1}\left(1-\dfrac{t^2}{3}\right)dt$$

$$I'=-\left(1-\frac{1}{9}\right)=-\frac{8}{9}$$

So the answer would be $\dfrac{1}{2}\left(\dfrac{-\pi\ln(2)}{2}-I'\right)=\dfrac{1}{2}\left(\dfrac{-\pi\ln(2)}{2}+\dfrac{8}{9}\right)$

But actual answer is $\dfrac{\pi}{8}\left(1-\ln4\right)$. What mistake am I making here?

Zacky
  • 30,116
user3290550
  • 3,580
  • so for $I'=\int_{0}^{\frac{\pi}{4}}\cos(2\theta)\ln(\tan\theta)d\theta$, how should we proceed. – user3290550 Jan 02 '20 at 20:03
  • Your approach is perfect, with the exception of that substitution $\tan \theta=t$, now you just need to continue. As a hint, integrate by parts. – Zacky Jan 02 '20 at 20:05
  • did you try beta function after setting $x^2=y$? – Ali Olaikhan Jan 02 '20 at 20:31
  • Moving this to the comments, since the integral appears on MSE, many times. Your mistake appears when you substitute $\tan \theta =t$, afterwards you should instead get: $$I'=\int_0^1 \frac{1-t^2}{(1+t^2)^2}\ln t,dt$$ Now notice that: $$\int \frac{1-t^2}{(1+t^2)^2}dt=-\int \frac{1-\frac{1}{t^2}}{\left(t+\frac{1}{t}\right)^2}dt=-\int \frac{\left(t+\frac{1}{t}\right)'}{\left(t+\frac{1}{t}\right)^2}dt=\frac{1}{t+\frac{1}{t}}+C$$ Thus we can integrate by parts to get: $$I'= \underbrace{\frac{\ln t}{t+\frac{1}{t}}\bigg|0^1}{=0} -\int_0^1 \frac{1}{1+t^2}dt=-\frac{\pi}{4}$$ – Zacky Jan 02 '20 at 20:34
  • See for example: https://math.stackexchange.com/q/1559380/515527 and https://math.stackexchange.com/q/1538682/515527. – Zacky Jan 02 '20 at 20:36

1 Answers1

1

Actually， \begin{align*} \int_0^{\frac\pi2}\cos(2\theta)\ln(\sin\theta)d\theta&= \frac12\int_0^{\frac\pi2}\ln(\sin\theta)d(\sin2\theta)\\ &=\frac12\sin2\theta\ln(\sin\theta)\bigg|_0^{\frac\pi2}-\frac12\int_0^{\frac\pi2}\sin2\theta\cdot\frac{\cos\theta}{\sin\theta}d\theta\\ &=-\frac12\int_0^{\frac\pi2}2\cos^2\theta d\theta=-\frac\pi4. \end{align*}

Misaya
  • 111