Monday, August 26, 2019

integration - Find limnrightarrowinftyintn0(1+fracxn)n+1exp(2x),dx




Find:



limnn0(1+xn)n+1exp(2x)dx



The sequence (1+xn)n+1exp(2x) converges pointwise to exp(x). So if we could apply Lebesgue's monotone convergence theorem, we have:



limnn0(1+xn)n+1exp(2x)dx=limnRI(0,n)(x)(1+xn)n+1exp(2x)dx=exp(x)dx=+



Can someone help me to prove that the sequence of integrands is monotone?


Answer



Using the indicator function, we can write



n0(1+xn)n+1e2xdx=0(1+xn)n+1e2xξ[0,n](x)dx



Then, note that for 0xn, we have




(1+xn)n+1=(1+xn)(1+xn)n2ex




EDIT: The OP requested a source for the preceding inequality.
In THIS ANSWER, I showed using only the limit definition of the exponential function along with Bernoulli's Inequality that the exponential function satisfies the inequality ex(1+xn)n

for x>n.




We can assert, therefore, that



(1+xn)n+1e2xξ[0,n](x)2ex




Applying the Dominated convergence theorem, we have



limnn0(1+xn)n+1e2xdx=0limn((1+xn)n+1ξ[0,n](x))e2xdx=0ex(1)e2xdx=1


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