# example of improper integral

The integrand of

 $\displaystyle I\;=\;\int_{0}^{1}\frac{\arctan{x}}{x\sqrt{1\!-\!x^{2}}}\,dx$ (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

 $\displaystyle I(y)\;=\;\int_{0}^{1}\frac{\arctan{xy}}{x\sqrt{1\!-\!x^{2}}}\,dx.$ (2)

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

 $f(x,\,y)\in O(1)\mbox{\; as \;}x\to 0,\quad f(x,\,y)\in O(\frac{1}{\sqrt{1\!-% \!x^{2}}})\mbox{\; as \;}x\to 1,$

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

 $\displaystyle\int_{0}^{1}\frac{\partial f(x,\,y)}{\partial y}\,dx$ (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)

 $x\;:=\;\cos\varphi,\quad\tan\varphi\;:=\;t$

which yield

 $\displaystyle I^{\prime}(y)$ $\displaystyle\;=\;\int_{0}^{1}\!\frac{dx}{(1\!+\!x^{2}y^{2})\sqrt{1\!-\!x^{2}}% }\;=\;\int_{0}^{\frac{\pi}{2}}\!\frac{d\varphi}{1\!+\!y^{2}\cos^{2}\varphi}$ $\displaystyle\;=\;\int_{0}^{\infty}\!\frac{dt}{1\!+\!y^{2}\!+\!t^{2}}\;=\;% \operatornamewithlimits{\Big{/}}_{\!\!\!t\,=0}^{\,\quad\infty}\!\frac{1}{\sqrt% {1\!+\!y^{2}}}\arctan\frac{t}{\sqrt{1\!+\!y^{2}}}$ $\displaystyle\;=\;\frac{\pi}{2\sqrt{1\!+\!y^{2}}}.$

Hence,

 $I(y)\;=\;\frac{\pi}{2}\int_{0}^{y}\frac{dy}{\sqrt{1\!+\!y^{2}}}\;=\;% \operatornamewithlimits{\Big{/}}_{\!\!\!0}^{\,\quad y}\!\ln(y+\sqrt{1\!+\!y^{2% }})$

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

 $\displaystyle\int_{0}^{1}\frac{\arctan{x}}{x\sqrt{1\!-\!x^{2}}}\,dx\;=\;\frac{% \pi}{2}\ln(1\!+\!\sqrt{2}).$ (4)
Title example of improper integral ExampleOfImproperIntegral 2014-11-07 11:47:42 2014-11-07 11:47:42 pahio (2872) pahio (2872) 6 pahio (2872) Example msc 40A10 SubstitutionNotation