Which holomorphic function is this the real part of? 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.
 A: 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.
A: 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. 
