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How can you show using Watson's lemma, that for some infinitely differentiable function $K(s)$ and $ kt \gg 1$ that

$$\int_0^tK(s) e^{-k(t-s)} \text{d}s\approx K(t)\int_{-\infty}^t e^{-k(t-s)} \text{d}s$$

Does one say, let us change variables to $u=t-s$

$$\int_0^tK(s) e^{-k(t-s)} \text{d}s = \int_t^0 K(t-u) e^{-ku} (-\text{d} u) \\ = \int_0^t K(t-u) e^{-ku} \text{d}u \approx \frac{K(t) \Gamma(1)}{k} $$ By Watson's lemma as $ kt \gg 1$ and $ k \gg 1$ (the $n=0$ term in the summation of Watson's lemma).

This clearly doesn't work, would someone be able to point out where I have gone wrong here and how I can produce the desired result?

EDIT

Something I have tried is

$$\int_0^t K(t-u) e^{-ku} \text{d}u = \int_0^\varepsilon K(t-u) e^{-ku} \text{d}u + O(e^{k\varepsilon})$$

where $1/k\ll \varepsilon \ll 1$. Letting $z=ku$ gives $$\frac{1}{k}\int_0^{\varepsilon k} K\left(t-\frac{z}{k}\right)e^{-z}\text{d}z \sim \frac{K(t)}{k}\int_0^{\infty}e^{-z}\text{d}z$$ as $\varepsilon k \gg 1$ and by making the approximation $K\left(t-\frac{z}{k}\right)\sim K(t)$ as $k\rightarrow\infty$.

Therefore as $z =k(t-s)$ $$\int_0^tK(s) e^{-k(t-s)} \text{d}s\approx K(t)\int_{-\infty}^t e^{-k(t-s)} \text{d}s$$

Question update

The above method gives the correct result, where then in my original attempt have I made a mistake and specifically what mistake was it?

Note, the definition for Watson's lemma which I used originally is found here https://en.wikipedia.org/wiki/Watson%27s_lemma

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  • $\begingroup$ Watson's lemma proper does not allow for the case that $t$ varies with $k$, so you will not be able to prove this by only using Watson's lemma. Further, it does not seem to be true without additional restrictions on $t$ or $K$. For example, if $t=k$ and $K(x) = e^{-x^2}$ then $$\int_0^t K(s) e^{-k(t-s)}\,ds \sim \sqrt{\pi} e^{-3k^2/4} \not\sim \frac{K(t)}{k} = K(t) \int_{-\infty}^{t} e^{-k(t-s)}\,ds.$$ $\endgroup$ Nov 17 '15 at 12:35
  • $\begingroup$ @AntonioVargas: Thank you for this, I think I need to loosen the restriction and allow $k\gg 1$. I will post an edit soon. $\endgroup$
    – Freeman
    Nov 17 '15 at 12:37
  • $\begingroup$ If $t = O(1)$ it might be true, I think. $\endgroup$ Nov 17 '15 at 12:40
  • $\begingroup$ @AntonioVargas: This approximation should become more accurate as $t$ increases for my purposes. See new update. $\endgroup$
    – Freeman
    Nov 17 '15 at 12:48
  • $\begingroup$ The approximation $K(t-z/k) \sim K(t)$ seems crucial, since it fails for the counterexample $K(x) = e^{-x^2}$ I mentioned earlier. If that property is characteristic of the functions $K$ you're interested in then your new proposed method may indeed form the base of a proof if you can make it rigorous. $\endgroup$ Nov 17 '15 at 12:55
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$$\int_0^t K(t-u) e^{-ku} \text{d}u = \int_0^\varepsilon K(t-u) e^{-ku} \text{d}u + O(e^{k\varepsilon})$$

where $1/k\ll \varepsilon \ll 1$. Letting $z=ku$ gives $$\frac{1}{k}\int_0^{\varepsilon k} K\left(t-\frac{z}{k}\right)e^{-z}\text{d}z \sim \frac{K(t)}{k}\int_0^{\infty}e^{-z}\text{d}z$$ as $\varepsilon k \gg 1$ and by making the approximation $K\left(t-\frac{z}{k}\right)\sim K(t)$ as $k\rightarrow\infty$.

Therefore as $z =k(t-s)$ $$\int_0^tK(s) e^{-k(t-s)} \text{d}s\approx K(t)\int_{-\infty}^t e^{-k(t-s)} \text{d}s$$

This is equivalent to the initial work you did you massive cretin.

Many thanks to Antonio Vargas in pointing this out.

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