Thursday, 26 October 2017

real analysis - Calculating limnrightarrowinftyint[0,infty)fracsin(ex)1+nx2,dx



I want to calculate the limit of following Lebesgue-integral:



limn[0,)sin(ex)1+nx2dx



Therefore I wanted to apply Lebesgue's dominated convergence theorem.
fn(x) is measurable and fn0 pointwise. Now it holds:




|sin(ex)1+nx2|11+x2:=g(x)



The improper integral over does converge. That means f is lebesgue integrable.
Therefore
limn[0,)sin(ex)1+nx2dx=[0,)limnsin(ex)1+nx2dx=0


Consider fn(0)=sin1 does not converge to 0. So I can't apply the theorem, can I ?


Answer



You are on the right track: apply Lebesgue's dominated convergence with g(x)=11+x2 which is Lebesgue integrable in [0,+). Since fn(x)=sin(ex)1+nx20 for all x>0 the sequence (fn)n converges to zero almost everywhere on [0,+), that's enough for dominated convergence, and we may conclude that the limit of 0fn(x)dx is zero.



Alternative way (without dominated convergence):

|[0,)sin(ex)1+nx2dx|+0|sin(ex)|1+nx2dx+0dx1+nx2=[arctan(nx)n]+0=π2n.


So, again, the limit as n is zero.


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...