Integration by parts yields divergent integral [duplicate]

$$I = \int_0^1\dfrac{t \ln t}{\sqrt{1-t^2}}\mathrm{dt}$$

Attempt:

Let $$t = \sin x$$

$$\implies dt = \cos x dx$$

$$\implies I = \displaystyle \int_0^{\pi/2} \dfrac{\sin x \ln (\sin x) \cos x dx}{\cos x}$$

$$\implies I = -\ln (\sin x) \cos x|_{0}^{\pi/2} + \displaystyle\int_0^{\pi/2}\dfrac{\cos^2 x}{\sin x} dx$$

The second integral can be "evaluated" using $$\cos^2 x = 1- \sin^2 x$$. But unfortunately, there are finally two divergent integrals.

$$\implies I = -\ln (\sin x) \cos x|_{0}^{\pi/2} + \displaystyle\int_0^{\pi/2}\csc x dx$$ + $$\displaystyle\int_0^{\pi/2}\sin x dx$$

How do I handle this?

marked as duplicate by Namaste integration StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 15 '18 at 10:10

• Substitute $x=t^2$ in the original integral and integrate by parts. – Kemono Chen Oct 15 '18 at 5:21
• Exactly as the previous comment says, no need for the cos change of vars. – Hashimoto Oct 15 '18 at 5:27
• I think the integral is convergent. – kmitov Oct 15 '18 at 5:42
• The integral is divergent. BTW, actually $I=\lim_{a\to0}-\ln\sin x\cos x|_a^{\pi/2}+\int_a^{\pi/2}\csc xdx+\int_0^{\pi/2}\sin xdx$ – Kemono Chen Oct 15 '18 at 5:45

If you want to avoid divergent integrals, use $$1-\cos x$$ as the antiderivative of $$\sin x$$. $$I=\ln\sin x(1-\cos x)\Big|_0^{\pi/2}+\int_0^{\pi/2}(\cos x-1)\cot xdx\\ =0+\ln2-1=\ln2-1.$$

• How can we use 1- cos x as the antiderivative – Abcd Oct 15 '18 at 6:25
• Note that you used $\int_0^{\pi/2} \sin x\ln\sin xdx=\int\sin xdx \ln\sin x|_0^{\pi/2}-\int_0^{\pi/2}(\int \sin xdx)(\ln \sin x)'dx$. $\int \sin xdx=C-\cos x$. You should choose a suitable $C$ to avoid divergency. – Kemono Chen Oct 15 '18 at 6:29
• I have never seen integration being done that way. – Abcd Oct 15 '18 at 6:30
• It is just a fancy notation of integrating by parts. – Kemono Chen Oct 15 '18 at 6:32
• What method did you use to evaluate the second integral? – Abcd Oct 15 '18 at 6:48

I would just focus on the antiderivative.

Using you substitution and continuing with the tangent half-angle substitution, we should get $$I=\int \dfrac{\sin (x) \ln (\sin (x)) \cos(x) }{\cos x}\,dx=\int\sin (x) \ln (\sin (x))\,dx$$ and integration by parts would lead to $$I=\cos (x)+\log \left(\sin \left(\frac{x}{2}\right)\right)-\log \left(\cos \left(\frac{x}{2}\right)\right)-\cos (x) \log (\sin (x))$$ Now, looking at the Taylor expansion around $$x=0$$, $$I=(1-\log (2))+x^2 \left(\frac{\log (x)}{2}-\frac{1}{4}\right)+O\left(x^4\right)$$ and around $$x=\frac \pi 2$$ $$I=-\frac{1}{6} \left(x-\frac{\pi }{2}\right)^3+O\left(\left(x-\frac{\pi }{2}\right)^4\right)$$ and then the result already given by @Kemono Chen.

Edit

Back to $$t$$, $$\int\dfrac{t \ln t}{\sqrt{1-t^2}}\,{dt}=\sqrt{1-t^2} (1-\log (t))+\log \left(\frac{t}{1+\sqrt{1-t^2}}\right)$$