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I need to calculate the asymptotics of the integral $$\left(\int_0^1 \mathrm e^{-tx} f(t)\right)^j$$ for $x\to\infty$. I suspect (and would like to prove), that this behaves like $$\left(\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{x^{n+1}}\right)^j,$$ but I'm not sure how. I get this result if I give each of the integrals a different parameter $x_i$, use Watson's lemma on each of them seperatly (which I can because the asymptotic integral is finite for finite $x_i$) and then evaluate at $x_1 = \ldots = x_j$. Does that make sense?

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The exponent $j$ seems to be extraneous, so let's just consider the asymptotic behavior of $$ \int_0^1e^{-tx}f(t)\,\mathrm{d}t $$ Using the binomial theorem, we get $$ \begin{align} \int_x^\infty e^{-t}t^n\,\mathrm{d}t &=e^{-x}\int_0^\infty (t+x)^n\,e^{-t}\,\mathrm{d}t\\ &=e^{-x}\int_0^\infty\sum_{k=0}^n\binom{n}{k}t^{n-k}x^k\,e^{-t}\,\mathrm{d}t\\ &=e^{-x}\sum_{k=0}^n\binom{n}{k}(n-k)!x^k\\ &=e^{-x}\sum_{k=0}^n\frac{n!}{k!}x^k\\ &=O(x^ne^{-x}) \end{align} $$ Thus, $$ \begin{align} &\left|\int_0^1e^{-tx}f(t)\,\mathrm{d}t-\sum_{n=0}^{N-1}\frac{f^{(n)}(0)}{x^{n+1}}\right|\\ &=\frac1x\left|\int_0^xe^{-t}f(t/x)\,\mathrm{d}t-\sum_{n=0}^{N-1}\frac{f^{(n)}(0)}{x^n}\right|\\ &=\frac1x\left|\int_0^x e^{-t}\left(\sum_{n=0}^{N-1}\frac{t^nf^{(n)}(0)}{x^nn!}+O\left(\frac{t^N}{x^N}\right)\right)\,\mathrm{d}t-\sum_{n=0}^{N-1}\frac{f^{(n)}(0)}{x^n}\right|\\ &=\frac1x\left|\int_0^\infty e^{-t}\left(\sum_{n=0}^{N-1}\frac{t^nf^{(n)}(0)}{x^nn!}+O\left(\frac{t^N}{x^N}\right)\right)\,\mathrm{d}t-\sum_{n=0}^{N-1}\frac{f^{(n)}(0)}{x^n}\right.\\ &\hphantom{=\frac1x}\left.-\int_x^\infty e^{-t}\left(\sum_{n=0}^{N-1}\frac{t^nf^{(n)}(0)}{x^nn!}+O\left(\frac{t^N}{x^N}\right)\right)\,\mathrm{d}t\right|\\ &=\frac1x\left|\int_0^\infty e^{-t}O\left(\frac{t^N}{x^N}\right)\,\mathrm{d}t-\int_x^\infty e^{-t}\left(\sum_{n=0}^{N-1}\frac{t^nf^{(n)}(0)}{x^nn!}+O\left(\frac{t^N}{x^N}\right)\right)\,\mathrm{d}t\right|\\ &=O\left(\frac1{x^{N+1}}\right)+O\left(\tfrac1xe^{-x}\right) \end{align} $$ Therefore, $$ \int_0^1e^{-tx}f(t)\,\mathrm{d}t=\sum_{n=0}^{N-1}\frac{f^{(n)}(0)}{x^{n+1}}+O\left(\frac1{x^{N+1}}\right) $$ Then we can use the binomial theorem to get $$ \left(\int_0^1e^{-tx}f(t)\,\mathrm{d}t\right)^j=\left(\sum_{n=0}^{N-1}\frac{f^{(n)}(0)}{x^{n+1}}\right)^j+O\left(\frac1{x^{N+j}}\right) $$

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Your proof is for Watson's lemma, that one I have already done. I was just wondering how to justify the last step that seems to be obvious from the partial expansion. – filmor Oct 19 '12 at 8:03

You can use Laplace's method.

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Laplace's method is just a generalised form of Watson's lemma, which I already use. Thanks anyways for the link :) – filmor Oct 19 '12 at 8:04

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