Let $n$ be a positive integer and $p>0$. Define $$ f_{n,p}(t) = \frac{(p \ln t - 1) (\sin^{2n} t)}{t^{p+1}}, $$ and I was trying to look at the behaviour of $$ \sqrt{n}\lim_{p\to 0^+}\int_0^\infty f_{n,p}(t)dt $$ when $n\to\infty$.
I evaluated the following three integrals: $$ \sqrt{n}\int_\pi^\infty f_{n,p}(t)dt,\quad \sqrt{n}\int_0^\infty f_{n,p}(t)dt,\quad \sqrt{n}\int_0^\pi f_{n,p}(t)dt $$ for some specific value of large $n$ and small $p>0$ and got the following results in MMA 12.1.1 (version June 9, 2020):
With[{n = 20000, p = 0.0000001}, NIntegrate[Sqrt[n] (p Log[t] - 1) Sin[t]^(2 n)/t^(p + 1),
{t, Pi, Infinity}, MaxRecursion -> 40]]
-0.376124345064075
With[{n = 20000, p = 0.0000001}, NIntegrate[Sqrt[n] (p Log[t] - 1) Sin[t]^(2 n)/t^(p + 1),
{t, 0, Infinity}, MaxRecursion -> 40]]
-1.12838344570119
With[{n = 20000, p = 0.0000001}, NIntegrate[Sqrt[n] (p Log[t] - 1) Sin[t]^(2 n)/t^(p + 1),
{t, 0, Pi}, MaxRecursion -> 40]]
-1.12838344570105
The sum of the first and the third integral should be the second, so something is wrong here.
The third integral looks correct because I can manually calculate that $$ \lim_{n\to\infty} \lim_{p\to 0^+} \sqrt{n}\int_0^\pi f_{n,p}(t)dt = -\frac{2}{\sqrt{\pi}} = -1.128379\cdots, $$ which agrees with the numerical integration result. The second integral is probably wrong and I have no idea of what the correct value of the first integral should look like but I think it should be positive.
Is there some workaround to obtain at least consistent results for the three integrals?
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Background of the problem. The problem is related to https://math.stackexchange.com/questions/2827591/alternate-proof-for-weighted-alternating-shifted-central-binomial-sum-relation Instead of $k^s$ in the sum, it is $\ln k$ in my problem, that is, I was looking at the sum $$ S_n = \sum_{k=1}^n (-1)^k \binom{2n}{n+k}\ln k. $$ I wish to show that $S_n / 2^{2n} >c/\sqrt{n}$ for some constant $c>0$ and all sufficiently large $n$. Following a similar approach to the answer in the link above, this boils down to showing that $$ \lim_{p\to 0^+} p \int_0^\infty f_{n,p}(t) dt > -\frac{c}{\sqrt n} $$ for some positive constant $c < \frac{1}{\sqrt{\pi}}(\gamma+\ln 2)\approx 0.7167$. I have managed to prove this analytically.
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For a given value of $n$, Mathematica is able to evaluate the integral analytically (the answer is complicated, involving gamma and polygamma functions). So I did
N[Table[Sqrt[n]
Limit[Integrate[
Sin[t]^(2 n)/t^(1 + p) (p Log[t] - 1), {t, 0, Infinity},
Assumptions -> 0 < p < 1], p -> 0], {n, 10}]]
{-0.635181422730739, -0.551179369011369, -0.521962044454548, \ -0.507143476307785, -0.498189412540769, -0.492195004661885, \ -0.487901432379977, -0.484674935869789, -0.482161769590284, \ -0.480148962175569}
which makes me think what I wanted to show is true.
With[{n = 20000, p = 0.0000001}, Solve[(p Log[t] - 1) == 0, t]]
. You might ponder the size of the that abscissa and how to integrate the spikes that occur everyPi
along thet
axis. You'd need millions of digits of precision. $\endgroup$