Wednesday, 25 December 2013

calculus - Compute limlimitsxto+inftydfraclnxintx0frac|sint|trmdt.





Compute

limx+lnxx0|sint|tdt.





Maybe, we can solve it by L'Hospital's rule, but there still exists a difficulty here. Though x+ implies lnx+, we do not know the limit of the denominator. How to solve it?


Answer



Thanks to @Alex B.'s hint , I complete the solution. Please correct me if I'm wrong.



For any x>0, we can choose some nN such that nπx<(n+1)π. Thus, we obtain
nπ0|sint|tdtx0|sint|tdt<(n+1)π0|sint|tdt.




On one hand, notice that
nπ0|sint|tdt=π0|sint|tdt+n1k=1(k+1)πkπ|sint|tdt>n1k=1(k+1)πkπ|sint|tdt>n1k=1(k+1)πkπ|sint|(k+1)πdt=n1k=1(k+1)πkπ|sint|dt(k+1)π=2πnk=21k.




On the other hand, likewise,
(n+1)π0|sint|tdt=π0|sint|tdt+nk=1(k+1)πkπ|sint|tdt<π0dt+nk=1(k+1)πkπ|sint|kπdt=π+nk=1(k+1)πkπ|sint|dtkπ=π+2πnk=11k.



Therefore
2πnk=21k<x0|sint|tdt<2+2πnk=11k.




Since
lnnπlnx<ln(n+1)π,


we have
lnnππ+2πnk=11k<lnxx0|sint|tdt<ln(n+1)π2πnk=21k.



Applying the subsitution as follows
nk=11k=lnn+γ+εn,


(in fact, we only need to recall that nk=11k and lnn are equivalent infinities), we can readily infer that the limits of the both sides in the last expression are both equal to π2 under the process n(i.e. x+). Hence, according to the squeeze theorem, we can conclude that
lnxx0|sint|tdtπ2(x+),
which is what we want to evaluate.



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