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For integer $n\geq 0$, consider the integral $$I_n := 2\int_{-\infty}^{\infty}\cos^n(x)\sin(x)xe^{-x^2}dx.$$

Show that each $I_n>0$ and $I_{n+1}< I_n$.

I believe this should be true based on computing values of $n$ in Mathematica, but I do not know how to prove this. For $n=0$, it's straightforward from integration by parts and using the Fourier transform of $e^{-x^2}$. For $n=1$, one can write $\cos(x)\sin(x)=\frac12\sin(2x)$ and use integration by parts again plus the the Fourier transform of $e^{-x^2}$.

If it helps, one can use integration by parts and trig identities to rewrite

$$I_n = (n+1)\int_{-\infty}^{\infty}\cos^{n+1}(x)e^{-x^2}dx - n\int_{-\infty}^{\infty}\cos^{n-1}(x)e^{-x^2}dx. $$

Nik Quine
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    I don't know if it's intended by whoever asked the question but maybe you could use the closed form of the integrals from your last line: Evaluate: $\int_{0}^\infty e^{-x^2} \cos^n(x) dx $ (the integrand is even so $\int_{_\infty}^\infty = 2\int_0^\infty$ here) – Bruno B Aug 23 '23 at 13:04
  • $I_{2k} > 0$ is easy. – River Li Aug 23 '23 at 15:33
  • @RiverLi Sorry for bothering you, would it be possible for you to elaborate? Those integrals are all fairly close to $0$, so a formal proof of positivity would be nice, especially considering that for all $n$ the integrand takes negative values on non-negligeable sets, and I don't see a (simple) change of variables which would make the integrand positive. (I'll note though that decreasingness is probably more important: by the dominated convergence theorem $I_n$ tends to $0$, hence decreasingness + convergence to $0$ would automatically give positivity. But something is better than nothing!) – Bruno B Aug 27 '23 at 13:35
  • @BrunoB I don't have a full solution now. – River Li Aug 27 '23 at 15:12
  • @BrunoB This question wasn't posed to me as a problem. It arose as part of a computation in statistical mechanics. – Nik Quine Aug 31 '23 at 21:54
  • What I wrote in my edit is, from far away, the best way to get the solution by recursivity. – Claude Leibovici Sep 01 '23 at 12:24
  • @ClaudeLeibovici Would you elaborate on why your recursive identity gives the solution to the question? Or are you just suggesting that it might be helpful for obtaining a solution? – Nik Quine Sep 03 '23 at 14:27
  • I wrote For the pleasure of working an interesting family of integrals. I never said that it was an answer to the question. I had fun and I hank you for that. Cheers :-) – Claude Leibovici Sep 03 '23 at 14:39

1 Answers1

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For the pleasure of working an interesting family of integrals

$$I_n = 2\int_{-\infty}^{+\infty}\cos^n(x)\sin(x)\,x\,e^{-x^2}\,dx$$

Using Euler representation of the sine and cosine functions and the binomial expansion, for $n>1$, they write $$I_n=\sqrt{\pi }\,\, 2^{-n}\, e^{-\frac{(n+1)^2}{4} }\,\Big((n+1)+e^n \left((n-1)^2+T_n\right)\Big)$$

Alas, no clear pattern seems to appear.

the first $T_n$ being $$\left( \begin{array}{cc} n & T_n \\ 2 & 0 \\ 3 & 0 \\ 4 & 2 e^2 \\ 5 & 10 e^3 \\ 6 & 27 e^4+5 e^6 \\ 7 & 56 e^5+28 e^8 \\ 8 & 100 e^6+84 e^{10}+14 e^{12} \\ 9 & 162 e^7+192 e^{12}+84 e^{15} \\ 10 & 245 e^8+375 e^{14}+270 e^{18}+42 e^{20} \\ 11 & 352 e^9+660 e^{16}+660 e^{21}+264 e^{24} \\ 12 & 486 e^{10}+1078 e^{18}+1375 e^{24}+891 e^{28}+132 e^{30} \\ \end{array} \right)$$

Edit

What could also be done is to use the linearization of $\cos^n(x)$ and use

$$J_{2k}=2\int_{-\infty}^{+\infty}\cos(2kx)\sin(x)\,x\,e^{-x^2}\,dx$$ $$J_{2k}=\sqrt{\pi } e^{-k^2-\frac{1}{4}} (\cosh (k)-2 k \sinh (k))$$

$$J_{2k+1}=2\int_{-\infty}^{+\infty}\cos((2k+1)x)\sin(x)\,x\,e^{-x^2}\,dx$$

$$J_{2k+1}=\sqrt{\pi } e^{-(k+1)^2} \left(k+1-ke^{2 k+1} \right)$$

Update

If $n$ is even $$\cos^n(x)=\frac 1{2^{n-1}}\sum_{k=0}^{\frac n2}\binom{n}{k} \cos ( (2 k-n)x)+\frac 1{2^{n}}\binom{n}{\frac{n}{2}}$$

If $n$ is odd $$\cos^n(x)=\frac 1{2^{n-1}}\sum_{k=0}^{\frac {n-1}2}\binom{n}{k} \cos ( (2 k-n)x)$$

Claude Leibovici
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    This might lead to nothing, but there's some order in this chaos: looking at these $T_n$s, there is a striking correspondance with this sequence in the OEIS: https://oeis.org/A298637 (found through this: https://oeis.org/search?q=352%2C+660%2C+660&language=english&go=Search). Precisely, it seems like, taking the triangular array given, the $T_n$'s coefficients are just the array to which we substract the first two columns, and the second column is $(n-1)^2$ ($(n-2)^2$ in the table but your $T_n$ corresponds to their $T(n+1,k)$s) which appears in your expression too. Just in case that helps. – Bruno B Aug 24 '23 at 13:41
  • Thanks a lot, Claude, for your contribution. Do you know a priori that $T_n>0$? – Nik Quine Aug 24 '23 at 20:29