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 fn→0 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,∞)limn→∞sin(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+nx2→0 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=π2√n.
So, again, the limit as n→∞ is zero.
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