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I am looking for an example of a function $f$ such that $\lim_{t\to x_n}f(t)=\infty$ for infinitely many points $x_n$ and for which the Laplace transform $\mathscr{L}(f)$ exists. I am sure it must be elementary...

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up vote 2 down vote accepted

Try $f:(0,+\infty)\to(0,+\infty)$ with period $1$ and such that, for every $0\lt x\lt 1$, $$ f(x)=\frac1{\sqrt{x(1-x)}}. $$ Then $f(x)\to+\infty$ when $x\to n$, for every nonnegative integer $n$ and, for every $\lambda\gt0$, $$ \mathscr{L}(f)(\lambda)=\int_0^{+\infty}\mathrm e^{-\lambda x}f(x)\mathrm dx=\int_0^1\mathrm e^{-\lambda x}f(x)\mathrm dx\cdot\sum_{n\geqslant0}\mathrm e^{-\lambda n}, $$ hence $$ \mathscr{L}(f)(\lambda)=\frac1{1-\mathrm e^{-\lambda}}\int_0^1\frac{\mathrm e^{-\lambda x}}{\sqrt{x(1-x)}}\mathrm dx, $$ and the last integral converges for every $\lambda$ hence $\mathscr{L}(f)(\lambda)$ converges.

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Did you mean to write $\sum_{n\geqslant 0}e^{-\lambda n}$? – Eckhard Jan 14 '13 at 11:43
@Eckhard Yes. Thanks. Corrected. – Did Jan 14 '13 at 11:54

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