example of improper integral

The integrand of

$I={\displaystyle {\int }_0^1}{\displaystyle \frac{\arctan x}{x\sqrt{1-x^2}}}𝑑x$(1)

is undefined both at the lower and the upper limit.  However, the value of the improper integral exists and may be found via the more general integral

$I(y)={\displaystyle {\int }_0^1}{\displaystyle \frac{\arctan xy}{x\sqrt{1-x^2}}}𝑑x.$(2)

Denote the integrand of (2) by  $f(x,y)$.  For any fixed real value $y$,

$$f(x,y)\in O(1)\text{ as }x\to 0,f(x,y)\in O(\frac{1}{\sqrt{1-x^2}})\text{ as }x\to 1,$$

where the Landau big ordo (http://planetmath.org/formaldefinitionoflandaunotation) notation has been used.  Accordingly, the integral (2) converges for every $y$.

The inequality

$$\left|\frac{\partial f(x,y)}{\partial y}\right|=\frac{1}{(1+x^2{y}^2)\sqrt{1-x^2}}\leqq \frac{1}{\sqrt{1-x^2}}$$

and the convergence of the integral

$${\int }_0^1\frac{dx}{\sqrt{1-x^2}}=\frac{\pi }{2}$$

imply that the integral

${\displaystyle {\int }_0^1}{\displaystyle \frac{\partial f(x,y)}{\partial y}}𝑑x$(3)

http://planetmath.org/node/6277converges uniformly on the whole $y$-axis and equals $I^{\prime }(y)$.  For expressing this derivative in a closed form (http://planetmath.org/ExpressibleInClosedForm), one may utilise the changes of variable (http://planetmath.org/ChangeOfVariableInDefiniteIntegral)

$$x:=\cos \phi ,\tan \phi :=t$$

which yield

$I^{\prime }(y)$$\mathrm{\hspace{0.33em}}={\displaystyle {\int }_0^1}{\displaystyle \frac{dx}{(1+x^2{y}^2)\sqrt{1-x^2}}}={\displaystyle {\int }_0^{\frac{\pi }{2}}}{\displaystyle \frac{d\phi }{1+y^2{\cos }^2\phi }}$

$\mathrm{\hspace{0.33em}}={\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle \frac{dt}{1+y^2+t^2}}=\underset{t=0}{\overset{\mathrm{\infty }}{/}}{\displaystyle \frac{1}{\sqrt{1+y^2}}}\arctan {\displaystyle \frac{t}{\sqrt{1+y^2}}}$

$\mathrm{\hspace{0.33em}}={\displaystyle \frac{\pi }{2\sqrt{1+y^2}}}.$

Hence,

$$I(y)=\frac{\pi }{2}{\int }_0^y\frac{dy}{\sqrt{1+y^2}}=\underset{0}{\overset{y}{/}}\ln (y+\sqrt{1+y^2})$$

and the integral (1) equals  $I=I(1)=\frac{\pi }{2}\ln (1+\sqrt{2})$,  i.e.

${\displaystyle {\int }_0^1}{\displaystyle \frac{\arctan x}{x\sqrt{1-x^2}}}𝑑x={\displaystyle \frac{\pi }{2}}\ln (1+\sqrt{2}).$(4)