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I want to find a class of function $h(\tau)$ which makes the following limitation converges to zero.

$\lim_{t\rightarrow \infty }\int_{0}^{t}e^{-M\left( t-\tau \right) }h\left( \tau \right) d\tau=0$ with $M>0$.

One choice is to take $h(\tau)=e^{-N\tau}$ with $N>0$.

The question is that exponential function converges to zero too fast. Is there any other function can make the limitation converge to zero but has a slow convergent rate than exponential function. The function $h(\tau)$ should be continuous or piecewise continuous between $[0,\infty)$.

Thanks for any helps!

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up vote 1 down vote accepted

To see the limit like the following: $$ \mathfrak{L}(h) = \lim_{t\to \infty} e^{-Mt}\int^t_0 e^{M\tau}h(\tau)d\tau $$ and it is sorta related to some special functions, like exponential integral. Now for your question, there are indeed many choices of functions that bear a slower decay rate than $e^{-N\tau}$, for examples:

  • $\displaystyle h(\tau) = \frac{1}{\tau^s}$ for $s>0$, here be careful with the integral, for $s\geq 1$, modify or truncate the part near $\tau = 0$ to ease your work, similarly $\displaystyle h(\tau) = \frac{1}{(a+\tau)^s}$ or more generally $\displaystyle h(\tau) = \frac{1}{P(\tau)}$ where $P(\tau)$ is a polynomial of $\tau$ such that $\displaystyle \lim_{\tau\to \infty}P(\tau) = 0$.

  • $\displaystyle h(\tau) = \frac{1}{\ln(a+\tau)}$ for $a > 1$, for this one, you get even slower convergence rate as $t$ approaching $\infty$, by checking numerically using adaptive quadrature myself, this one converges really really slow.

  • $\displaystyle h(\tau) = \frac{\pi}{2} - \arctan (\tau)$, from the series expansion point of view, this one decays similar to an inverse of a polynomial.

  • $\displaystyle h(\tau) = \ln(1+\frac{1}{\tau})$, again this one decays in a similar rate with the first case.

Above choices would all make $\mathfrak{L}(h) = 0 $, to prove, integration by parts would give you something like $\mathfrak{L}(h')$, and you may wanna consider the case $\displaystyle h(\tau) = \frac{1}{\tau}$ first.

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