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  "body": "\n## Counterexample\n\nThe function $f(t) = e^{t^2}$ on $[0, \\infty)$ is continuous, but its [[Laplace integral|05XT5AVJBK]] $\\int_0^\\infty e^{t^2} e^{-st}\\, dt$ converges for no $s \\in \\C$.\nCheck: suppose it converged at some $s_0 \\in \\C$.\nBy [[the half-plane theorem|6X05N89XDP]] it would then converge at every real $s \u003e \\operatorname{Re} s_0$.\nBut for real $s$ and $t \\ge \\max(s, 0)$, the integrand satisfies $e^{t^2 - st} = e^{t(t - s)} \\ge 1$, so $\\int_0^T e^{t^2 - st}\\, dt \\to \\infty$ as $T \\to \\infty$.\nThis is a contradiction, so the abscissa of convergence of $f$ is $+\\infty$.\nThe function is not of [[exponential order|TKZAV3KR86]]: $e^{t^2} / e^{at} \\to \\infty$ for every real $a$.\n\n## Remarks\n\nExponential order is sufficient, not necessary, for a Laplace transform to exist.\nFor example, $f(t) = 2t e^{t^2} \\cos(e^{t^2})$ is not of exponential order, yet $\\int_0^T f(t)\\, dt = \\sin(e^{T^2}) - \\sin 1$ stays bounded, and integration by parts shows that its Laplace integral converges for every $s$ with $\\operatorname{Re} s \u003e 0$.\nA different obstruction is local: $f(t) = 1/t$ has no Laplace transform because $\\int_0^1 dt/t$ diverges, whatever $s$ is.\n\n## Explanations\n\n### L1 human\n\nThe Laplace transform weighs $f(t)$ by $e^{-st}$, which shrinks at most exponentially fast.\nThe function $e^{t^2}$ grows faster than any exponential, so no choice of $s$ can tame it and the integral always blows up.\nFast growth is not the only way to fail: a function such as $1/t$ is too large near $0$.\n",
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  "msc": [
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  "name": "Function with no Laplace transform",
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