Asymptotic behaviour of Convolution power

Question

We have a function $f(t): [0,\infty) \to \mathbb{R}$. The convolution of $f(t)$ with itself is: \begin{equation} (f*f)(t) = \int\limits_0^t \! \mathrm{d}\tau \; f(\tau) f(t-\tau) \end{equation}

We define $f^{*n}(t)$ as the convolution of $f$ with itself $n$ times, also called the $n$th convolution power. I would like to understand the the large $t$ asymptotic behaviour of the convolution power of $f(t)=\frac{\sinh(t)}{t}$: I think that $f^{*n}(t) \sim \frac{e^t}{t}$, but I can't find a way to prove it or disprove it. Also I'm not aware of general theorems regarding the asymptotic behaviour of convolution powers.

Answer 1

We have $$F(p) = \mathcal L[f](p) = \frac 1 2 \ln \frac {p + 1} {p - 1}, \\ \mathcal L [f^{*n}](p) =F^n(p).$$ Deforming the integration contour to run along the branch cut of the logarithm and applying Laplace's method, we get that the Bromwich integral is asymptotically equivalent to the sum of the integrals over the segments $[1 - \epsilon - i0, 1 - i0]$ and $[1 + i0, 1 - \epsilon + i0]$. Taking $p = 1 - \xi$ with real $\xi$, we obtain $$F^n(1 - \xi + i0) - F^n(1 - \xi - i0) = \left( \frac 1 2 \ln \left| \frac 2 \xi - 1 \right| - \frac {\pi i} 2 \right)^{\!n} - \left( \frac 1 2 \ln \left| \frac 2 \xi - 1 \right| + \frac {\pi i} 2 \right)^{\!n} \sim \\ -\pi n (-2)^{-n + 1} i \ln^{n - 1} \xi, \quad \xi \to 0^+, \\ \mathcal L^{-1}[F^n](t) \sim -n (-2)^{-n} \int_0^\infty e^{t (1 - \xi)} \ln^{n - 1} \xi \,d\xi = \\ -n (-2)^{-n} e^t \frac {d^{n - 1}} {da^{n - 1}} \int_0^\infty \xi^a e^{-t \xi} d\xi \,\bigg\rvert_{a = 0} = -n (-2)^{-n} e^t \frac {d^{n - 1}} {da^{n - 1}} (\Gamma(a + 1) t^{-a - 1}) \bigg\rvert_{a = 0} \sim \\ \frac {n e^t \ln^{n - 1} t} {2^n t}, \quad t \to \infty$$ (to get the highest power of $\ln t$, we need to differentiate the $t^{-a - 1}$ factor $n - 1$ times).

Answer 2

To start with those problems you can look at Laplace transform

$$F(s) = \int_0^\infty \frac{\sinh(t)}{t}e^{-st}dt, \qquad 2F'(s) = \int_0^\infty (e^{-t}-e^t)e^{-st}dt= \frac{1}{s+1}-\frac{1}{s-1},\\ 2 F(s) = \log(s+1)-\log(s-1)+F(2)-\log(2)= \log(1+\frac{2}{s-1})+F(2)-\log(2)$$

$e^{-2t} \frac{\sinh(t)}{t} $ is integrable so $\lim_{s \to +\infty} F(s+2) = 0$ and $$F(s)=\frac12 \log(1+\frac{2}{s-1}), \qquad F(s)^n = \sum_{m=n}^\infty c_{n,m } (s-1)^{-m}$$ Where the $c_{n,m}$ are the coefficients of $2^{-n}\log^n(1+2z)$ at $z=0$.

And since $\int_0^\infty t^{m-1}e^t e^{-st}dt = (s-1)^{-m}(m-1)!$ then

$$f^{*n}(t) =\mathcal{L}^{-1}[F(s)^n](t)=e^t\sum_{m=n}^\infty \frac{c_{n,m}}{(m-1)!} t^{m-1}$$