Compute
limx→+∞lnx∫x0|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 n∈N such that nπ≤x<(n+1)π. Thus, we obtain
∫nπ0|sint|tdt≤∫x0|sint|tdt<∫(n+1)π0|sint|tdt.
On one hand, notice that
∫nπ0|sint|tdt=∫π0|sint|tdt+n−1∑k=1∫(k+1)πkπ|sint|tdt>n−1∑k=1∫(k+1)πkπ|sint|tdt>n−1∑k=1∫(k+1)πkπ|sint|(k+1)πdt=n−1∑k=1∫(k+1)πkπ|sint|dt(k+1)π=2πn∑k=21k.
On the other hand, likewise,
∫(n+1)π0|sint|tdt=∫π0|sint|tdt+n∑k=1∫(k+1)πkπ|sint|tdt<∫π0dt+n∑k=1∫(k+1)πkπ|sint|kπdt=π+n∑k=1∫(k+1)πkπ|sint|dtkπ=π+2πn∑k=11k.
Therefore
2πn∑k=21k<∫x0|sint|tdt<2+2πn∑k=11k.
Since
lnnπ≤lnx<ln(n+1)π,
we have
lnnππ+2πn∑k=11k<lnx∫x0|sint|tdt<ln(n+1)π2πn∑k=21k.
Applying the subsitution as follows
n∑k=11k=lnn+γ+εn,
(in fact, we only need to recall that n∑k=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
lnx∫x0|sint|tdt→π2(x→+∞),
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