example of improper integral


The integrand of

I=∫01arctan⁡xx⁢1-x2⁢𝑑x (1)

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

I⁢(y)=∫01arctan⁡x⁢yx⁢1-x2⁢𝑑x. (2)

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

f⁢(x,y)∈O⁢(1)⁢ as ⁢x→0,f⁢(x,y)∈O⁢(11-x2)⁢ as ⁢x→1,

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

The inequalityMathworldPlanetmath

|∂⁡f⁢(x,y)∂⁡y|=1(1+x2⁢y2)⁢1-x2≦11-x2

and the convergence of the integral

∫01d⁢x1-x2=π2

imply that the integral

∫01∂⁡f⁢(x,y)∂⁡y⁢𝑑x (3)

http://planetmath.org/node/6277converges uniformly on the whole y-axis and equals I′⁢(y).  For expressing this derivative in a closed formMathworldPlanetmath (http://planetmath.org/ExpressibleInClosedForm), one may utilise the changes of variable (http://planetmath.org/ChangeOfVariableInDefiniteIntegral)

x:=cos⁡φ,tan⁡φ:=t

which yield

I′⁢(y)  =∫01d⁢x(1+x2⁢y2)⁢1-x2=∫0π2d⁢φ1+y2⁢cos2⁡φ
 =∫0∞d⁢t1+y2+t2=/t=0∞⁡11+y2⁢arctan⁡t1+y2
 =π2⁢1+y2.

Hence,

I⁢(y)=π2⁢∫0yd⁢y1+y2=/0y⁡ln⁡(y+1+y2)

and the integral (1) equals  I=I⁢(1)=π2⁢ln⁡(1+2),  i.e.

∫01arctan⁡xx⁢1-x2⁢𝑑x=π2⁢ln⁡(1+2). (4)
Title example of improper integral
Canonical name ExampleOfImproperIntegral
Date of creation 2014-11-07 11:47:42
Last modified on 2014-11-07 11:47:42
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 6
Author pahio (2872)
Entry type Example
Classification msc 40A10
Related topic SubstitutionNotation