# Asymptotic behaviour of Convolution power

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

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.

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).

• You are saying $f^{*n}(t) = \frac1{2i\pi} \int_{2+-i\infty}^{2+i\infty} F(s)^n e^{st} ds = \frac1{2i\pi} \int_C F(s)^n e^{st} ds$ where $C$ encloses $[-1,1]$ and for $t$ large it is $\sim \frac1{2i\pi} \int_{1-a}^1 F(s)^ne^{st}ds+\frac1{2i\pi}\int_1^{1-a} (F(s)+i\pi)^n e^{2i\pi})e^{st}ds$ – reuns Mar 18 at 6:04
• That is correct. – Maxim Mar 18 at 13:46
• Thank you very much for your answer! In the third line you extend tha integration upper limit to infinity because the asymptotic behaviour remains unaltered? – G.Carugno Mar 29 at 16:40
• @G.Carugno Correct, that's the idea of Laplace's method: we can contract the interval of integration to an arbitrarily small interval around the critical point without changing the asymptotic behavior, replace the integrand by a simpler one and extend the interval of integration indefinitely also without changing the asymptotic behavior. – Maxim Mar 29 at 18:34

$$F(s) = \int_0^\infty \frac{\sinh(t)}{t}e^{-st}dt, \qquad 2F'(s) = \int_0^\infty (e^{-t}-e^t)e^{-st}dt= \frac{1}{s+1}-\frac{1}{s-1},\\ 2 F(s) = \log(s+1)-\log(s-1)+F(2)-\log(2)= \log(1+\frac{2}{s-1})+F(2)-\log(2)$$
$$e^{-2t} \frac{\sinh(t)}{t}$$ 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
$$f^{*n}(t) =\mathcal{L}^{-1}[F(s)^n](t)=e^t\sum_{m=n}^\infty \frac{c_{n,m}}{(m-1)!} t^{m-1}$$
• Thanks for the useful insight! The radius of convergence of the series you used for $2^{-n} \log^n(1+2z)$ around 0 is $\frac{1}{2}$. Shouldn't it be a problem when you invert the Laplace transform? – G.Carugno Mar 16 at 20:08
• As you see no, the obtained $f^{*n}(t)$ is entire, a consequence of $G(z)=F(1/z)$ being analytic and $G(0) = 0$. $\Re(s) > 3$ implies $|(s-1)^{-1}| < 1/2$ so on the vertical line $\Re(s)=4$ the series in $(s-1)^{-1}$ converges in $L^1$ which is all you need to find $e^{4t} f^{*n}(t)$ – reuns Mar 16 at 20:17
• The last sum is not an asymptotic series for large $t$ though. – Maxim Mar 17 at 1:25