Is there a nice way to find this integral $\int_0^1\frac{ \arcsin x}{x} \mathrm{d}x$?

Question

$$\int_0^1\frac{ \arcsin x}{x}\,\mathrm dx$$

I was looking in my calculus text by chance when I saw this example , the solution is written also but it uses very tricky methods for me ! I wonder If there is a nice way to find this integral.

The idea of the solution in the text is in brief , Assume $y=\sin(x)$ and use definition of improper integral and some properties of definite integral to get $ -\lim_{\varepsilon \rightarrow0^+} \int_\epsilon^{\frac{\pi}{2}} \ln(\cos(y-\varepsilon))\,dy$ then it uses that fact that $a= (a+a) \times \frac{1}{2}$ where $a$ here is the integral , and then there is a step which I can't understand till the moment( but understand most of the rest of the steps ) . and it keeps going to use more and more tricks to get the final result , $\frac{\pi}{2} \ln2$.

Now, I try to understand this method , If I couldn't I will ask for help, but in the moment, I wonder if there is a good way to find this integral.

Answer 1

$$I = \int_0^1 \frac{\arcsin x}{x}\,dx = \int_{0}^{\pi/2}\frac{x\,dx}{\sin x}\cos x = x\log\sin x \bigg|_0^{\pi/2}-\int_0^{\pi/2} \log\sin x\,dx$$

This last integral is well known to equal $-\frac{\pi}{2}\log 2$. Thus

$$I = \frac{\pi}{2}\log 2$$

Answer 2

Using integration by parts with $u=\arcsin(x)$ yields

$$ \int_0^1 \frac{\arcsin x}{x}\,dx = -\int_0^1 \frac{\ln(x)}{\sqrt{1-x^2}}\,dx=I .$$

Now, consider the integral

$$ F = \int_0^1 \frac{x^\alpha}{\sqrt{1-x^2}}\,dx = \frac{\sqrt{\pi}}{2}{\frac {\Gamma \left( \frac{\alpha}{2}+\frac{1}{2} \right) }{ \Gamma \left( \frac{\alpha}{2}+1 \right) }},$$

which can be evaluated using the $\beta$ function (subs $x^2=t$ in F). Now, $I$ follows from $F$ as

$$ I = \lim_{\alpha\to 0}\frac{dF}{d\alpha}= \frac{\pi \ln(2)}{2}.$$

Answer 3

Let $$ y=sin^{-1}x\ \ \Rightarrow x= \sin y,\ \ dx= \cos y dy,$$ then $$ I=\int_{0}^{1}\frac{sin^{-1}x}{x}dx=\int_{0}^{\frac{\pi }{2}}\frac{y}{siny}.cosy=\int_{0}^{\frac{\pi }{2}}\frac{y}{tany}dy.$$ Let $$I(a)=\int_{0}^{\frac{\pi }{2}}\frac{\arctan(a \tan y)}{\tan y}dy \Rightarrow I'(a)=\int_{0}^{\frac{\pi }{2}}\frac{dy}{1+(a \tan(y))^2}.$$ One can easily prove that $$\int_{0}^{\frac{\pi }{2}}\frac{1}{1+(a \tan(y))^2}dy=\frac{\pi }{2(1+a)}.$$ Therefore $$I'(a)=\frac{\pi }{2(1+a)}\ \ \Rightarrow I(a)=\frac{\pi }{2}\ln(1+a)+c$$ and therefore $$I(a)=\int_{0}^{\frac{\pi }{2}}\frac{\arctan(a \tan x)}{\tan x}dx=\frac{\pi }{2}\ln(1+a)+c.$$ Substitute $a=0$ to find $c=0,$ therefore $$I(a)=\frac{\pi }{2}\ln(1+a).$$ Now putting $a=1,$ we get $$\int_{0}^{\frac{\pi }{2}}\frac{x}{\tan x}dx=\frac{\pi }{2}\ln(2).$$

Answer 4

Let $$I=\int_{0}^{1} \frac{\arcsin(x)}{x} dx.$$ Put $x=\sin(\theta), dx=\cos(\theta) d\theta$ so that $$I=\int_{0}^{\frac{\pi}{2}}\frac{\theta}{\tan(\theta)} d\theta.$$ We will show this representation of $I$ is indeed equal to a double integral that we can easily evaluate in two ways.

Consider $$J=\int_{0}^{1}\int_{0}^{\infty}\frac{1}{(1+x^2y^2)(1+y^2)} dydx.$$ Using partial fractions, you can integrate with respect to $y$ first as such:

$$\frac{1}{(1+x^2y^2)(1+y^2)}=\frac{1}{1-x^2} \left(\frac{-x^2}{1+x^2y^2}+\frac{1}{1+y^2}\right).$$ so $$J=\int_{0}^{1} \frac{1}{1-x^2} \left( -\frac{\pi}{2}x +\frac{\pi}{2} \right)dx=\int_{0}^{1} \frac{\pi}{2}\left(\frac{1}{1+x}\right) dx=\frac{\pi \ln(2)}{2}.$$ On the other hand, let us try to integrate $J$ the other away around as such: $$J=\int_{0}^{\infty}\int_{0}^{1}\frac{1}{(1+x^2y^2)(1+y^2)} dxdy.$$

$$J=\int_{0}^{\infty} \frac{\tan^{-1}(y)}{y(y^2+1)}dy.$$ Now let $y=\tan(\theta),dy=\sec^{2}(\theta) d\theta$ so that we get:

$$J=\int_{0}^{\frac{\pi}{2}}\frac{\theta}{\tan(\theta)} d\theta,$$ so we get that $$J=I=\frac{\pi \ln(2)}{2}.$$

Answer 5

By integration by parts $$ \begin{aligned} \int_0^1 \frac{\arcsin x}{x} d x & =[\ln x \arcsin x]_0^1-\int_0^1 \frac{\ln x}{\sqrt{1-x^2}} d x \\ & \\ & =-\int_0^{\frac{\pi}{2}} \ln (\sin \theta) d \theta \\ & \\ & =\frac{\pi}{2}\ln 2 \end{aligned} $$

Answer 6

The most efficient way to do this integral would probably be by transforming it into the log-sin integral, as mentioned in previous answers. However, another interesting way to solve the problem is by using the substitution $x=\sin(t)$ to obtain $$ \int_{0}^{1}\frac{\arcsin(x)}{x}dx=\int_{0}^{\pi/2}\frac{t\cos(t)}{\sin(t)}dt=\int_{0}^{\pi/2}\frac{t}{\tan(t)}dt $$ From here, consider the function $$ F(\alpha)=\int_{0}^{\pi/2}\frac{\tan^{-1}(\alpha\tan(t))}{\tan(t)}dt $$ Notice that your integral is equivalent to $F(1)$, as the tan and arctan cancel out. If we differentiate $F$ with respect to $\alpha$ we get $$ \frac{dF(\alpha)}{d\alpha}=\int_{0}^{\pi/2}\frac{\partial }{\partial \alpha}\frac{\tan^{-1}(\alpha\tan(t))}{\tan(t)}dt=\int_{0}^{\pi/2}\frac{1}{(1+\alpha^2\tan^2(t))}dt $$ This means that we consider our integrand as a function of $\alpha$ and differentiate it under the integral sign. To solve this integral, first multiply by $\sec^2(t)$ on the top and bottom, then use Pythagorean identity to obtain $$ \frac{dF}{d\alpha}=\int_{0}^{\pi/2}\frac{\sec^2(t)}{(1+\alpha^2\tan^2(t))(1+\tan^2(t))}dt $$ Substituting $u=\tan(t)$ leaves $$ \int_{0}^{\infty}\frac{1}{(1+\alpha^2u^2)(1+u^2)}du=\frac{1}{\alpha^2-1}\int_{0}^{\infty}\frac{\alpha^2}{1+\alpha^2u^2}-\frac{1}{1+u^2}du $$ This can be easily integrated to get $$ \frac{dF}{da}=\frac\pi2 \cdot\frac{1}{\alpha+1} $$ Recall that we want $F(1)$, so by FTC we have $$ F(\alpha)=\int_{0}^{\pi/2}\frac{\tan^{-1}(\alpha\tan(t))}{\tan(t)}dt =\frac{\pi}{2}\log(\alpha+1)+C $$ When $\alpha=0$ the integral is $0$, so we have $$ F(0)=0\implies C=0 $$ Finally, our answer becomes $$ \boxed{\frac{\pi}{2}\log(2)} $$

Answer 7

I think what the questioner is asking about is (i) how to deal with the fact that the integral is improper and (ii) the tricks used to integrate $\int_0^{\frac{1}{2}\pi}\ln(\sin x) dx$. On the latter, here is a possible sequence.

By substitution of $y=\frac{1}{2}\pi - x$, you get $$\int_0^{\frac{1}{2}\pi}\ln(\sin x) dx =\int_0^{\frac{1}{2}\pi}\ln(\cos y) dy = \int_0^{\frac{1}{2}\pi}\ln(\cos x) dx .$$ Therefore $$\int_0^{\frac{1}{2}\pi}\ln(\sin x) dx =\frac{1}{2}\left(\int_0^{\frac{1}{2}\pi}\ln(\sin x) dx + \int_0^{\frac{1}{2}\pi}\ln(\cos x) dx\right) =\frac{1}{2}\int_0^{\frac{1}{2}\pi}\ln(\sin x \cos x) dx.$$ So $$\int_0^{\frac{1}{2}\pi}\ln(\sin x) dx =\frac{1}{2}\int_0^{\frac{1}{2}\pi}\ln(\tfrac{1}{2}\sin 2x) dx =-\frac{1}{2}\int_0^{\frac{1}{2}\pi}\ln 2 \;dx+\frac{1}{2}\int_0^{\frac{1}{2}\pi}\ln(\sin 2x) dx.$$ But, by substitution of $y=2x$, you get $$\int_0^{\frac{1}{2}\pi}\ln(\sin 2x) dx =\frac{1}{2}\int_0^\pi\ln(\sin y) dy = \frac{1}{2}\int_0^\pi\ln(\sin x) dx = \int_0^{\frac{1}{2}\pi}\ln(\sin x) dx ,$$ where the last step comes from symmetry of the $\sin$ function about $\frac{1}{2}\pi$.

Then $$\int_0^{\frac{1}{2}\pi}\ln(\sin x) dx =-\tfrac{1}{4}\pi \ln 2+\frac{1}{2}\int_0^{\frac{1}{2}\pi}\ln(\sin x) dx$$ and so, finally, $$\int_0^{\frac{1}{2}\pi}\ln(\sin x) dx =-\tfrac{1}{2}\pi \ln 2.$$