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In the paper "The Inverse Function Theorem of Nash and Moser" by Richard S. Hamilton it is claimed that there exists a function $\phi$ such that:

$$\int_{0}^{\infty}t^{n}\phi(t)dt=(-1)^{n}$$

For $n=0,1,2,...$. In fact one is provided. The example is:

$$\phi(t)=\frac{e^{2\sqrt{2}}}{\pi(1+t)}e^{-(t^\frac{1}{4}+t^\frac{-1}{4})}\sin(t^\frac{1}{4}-t^{\frac{-1}{4}})$$

My trouble is that the paper said that verifying the integral can be done by methods of contour integration by recognizing this function as the real part of a holomorphic function. The trouble I'm having is that I have no idea how to identify this holomorphic function. Any help is greatly appreciated. Thank you in advance.

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This reminds me of the function used in E.M. Stein's book on singular integrals being $\psi(t) = \frac{e}{\pi t} e^{-\sqrt{2}/2 (t-1)^{1/4}} sin(\sqrt{2}/2 (t-1)^{1/4}$, which satisfies $\int_1^\infty \psi(t) dt = 1$ and all higher moments vanish. –  andre Aug 4 at 22:03
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$\psi(t) = \frac{e}{\pi t} e^{-\sqrt{2}/2 (t-1)^{1/4}} sin(\sqrt{2}/2 (t-1)^{1/4})$ –  andre Aug 4 at 22:11
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And also the one in D.V. Widder's book on the laplace transform, being $\phi(t) = e^{-t^{1/4}}sin(t^{1/4})$, where all the moments vanish for the integral on $[0,\infty)$, see archive.org/stream/laplacetransform031816mbp#page/n141/mode/2up –  andre Aug 12 at 20:47
    
And as stated in D.V. Widders book the last $\phi$ can be found in a 1894 paper of Stieltjes on page J. 105, see archive.numdam.org/ARCHIVE/AFST/AFST_1894_1_8_4/…, I wonder how Stieltjes found this function. –  andre Aug 13 at 22:53

2 Answers 2

up vote 6 down vote accepted

Wahoo, this is cool. Replace $t$ with $u^4$ in order to have: $$I_n=\int_{0}^{+\infty}t^n \phi(t)\,dt = \frac{4e^{2\sqrt{2}}}{\pi}\int_{0}^{+\infty}\frac{u^{4n+3}}{1+u^4}\sin(u-1/u)\exp(-u-1/u)\,du,$$ then split $[0,+\infty)=[0,1]\cup[1,+\infty)$ and use the substitution $u\leftarrow 1/u$ on the second interval in order to have: $$I_n = \frac{4e^{2\sqrt{2}}}{\pi}\int_{0}^{1}\frac{u^{4n+2}-u^{-4n-2}}{u^2+u^{-2}}\sin(u-1/u)\exp(-u-1/u)\frac{du}{u}.$$ Now substitute $u=e^{-v}$ in order to have: $$I_n = \frac{4e^{2\sqrt{2}}}{\pi}\int_{0}^{+\infty}\frac{\sinh((4n+2)v)}{\cosh(2v)}\sin(2\sinh v)\exp(-2\cosh v)\,dv,$$ and since the integrand function is even: $$I_n = \frac{2e^{2\sqrt{2}}}{\pi}\Im\int_{\mathbb{R}}\frac{\sinh((4n+2)v)}{\cosh(2v)}\exp(2i\sinh v-2\cosh v)\,dv.$$ We can approach the last integral with the usual complex analytic techniques, by integrating the function over a rectangle having vertices in $-R,R,R+\frac{\pi}{2}i,-R+\frac{\pi}{2}i$. The only singularity that matters is the one in $v=\frac{\pi}{4}i$: since the residue of the integrand function in such a point is: $$\frac{(-1)^n}{2}e^{-2\sqrt{2}}$$ we have $I_n=(-1)^n$ just as claimed.

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I don't mean to ruin your masterpiece but I think the substitution at the beginning: $t=u^{4}$ results in the factor of $u^{4n+3}$ within the integral. Sorry for bothering you but I'm a neurotic raccoon. In any case great answer. (+1) –  user71352 Aug 4 at 4:39
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@user71352: you are right, just a typo. Many thanks. –  Jack D'Aurizio Aug 4 at 4:42
    
You're welcome. –  user71352 Aug 4 at 4:43
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@Jack D'Aurizio Thank you very much for your help! –  user99163 Aug 4 at 4:46

In general, an analytic function must have harmonic real and imaginary parts (i.e. they satisfy Laplace's equation) and satisfy the Cauchy-Riemann equations.

In this case, replace the sine into a complex exponential.

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