The Laplace transform is defined as an integral from 0 to \infty — an improper integral by its very construction. Whether that transform even exists for a given function comes down to exactly the convergence question this section reviews.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus II — Integral Calculus & Infinite Series
Find the value of k (as a decimal) such that \displaystyle\int_1^\infty \frac{1}{x^k}\,dx = \frac{1}{3}.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
\int_1^\infty x^{-k}dx = \frac1{k-1} for k>1; setting \frac1{k-1}=\frac13 gives k-1=3, so k=4.
Correct answer
Also accepted:
Why
The value of the convergent integral is \frac1{k-1} for k>1.
From Calculus II — Integral Calculus & Infinite Series
\displaystyle\int_0^\infty e^{-x}\,dx
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The limit of the antiderivative is not zero -- it settles at a positive finite value.
The limit DOES exist and is finite -- this integral is convergent, not divergent.
e isn't the value that results from evaluating this specific limit.
Why
\lim_{t\to\infty}\left[-e^{-x}\right]_0^t = \lim_{t\to\infty}(-e^{-t}+1) = 0+1 = 1.
Replace \infty with a finite t, find an antiderivative of e^{-x}, evaluate at t and 0, then take the limit as t\to\infty.
From Calculus II — Integral Calculus & Infinite Series
An improper integral \int_a^\infty f(x)\,dx is defined as:
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"f(\infty)" isn't a meaningful expression -- infinity isn't a number you evaluate a function at.
Plugging \infty directly into an antiderivative is exactly the illegal operation this definition avoids.
It has a precise, rigorous definition -- as a limit of ordinary definite integrals.
Why
The infinite bound is always handled by replacing it with a finite variable and taking a limit, never by direct substitution.
An infinite bound is never plugged in directly -- what replaces it before a limit is taken?
An improper integral asks what happens to an ordinary area calculation as the far edge is pushed out to infinity. Sometimes that area settles down to a fixed number; sometimes it keeps growing forever. The whole question is which of those two things happens.
An integral with an infinite bound is defined as a limit of an ordinary definite integral: \displaystyle\int_a^\infty f(x)\,dx = \lim_{b\to\infty}\int_a^b f(x)\,dx. The improper integral converges if that limit exists as a finite number and diverges otherwise. The Laplace transform \displaystyle\int_0^\infty f(t)e^{-st}\,dt is exactly this kind of integral, and it converges precisely when the decaying factor e^{-st} beats the growth rate of f(t) as t\to\infty — which is why the transform is only guaranteed to exist for s larger than some threshold tied to how fast f itself grows.
Watch the area settle as b grows
Drag b outward. Watch whether the shaded area under 1/x^2 approaches a fixed number or keeps growing.
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Worked example
A convergent improper integral
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1
Evaluate \displaystyle\int_0^\infty e^{-x}\,dx.
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2
Write it as a limit: \displaystyle\lim_{b\to\infty}\int_0^b e^{-x}\,dx = \lim_{b\to\infty}\big[-e^{-x}\big]_0^b = \lim_{b\to\infty}(1-e^{-b}).
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3
As b\to\infty, e^{-b}\to0, so the limit is 1. \boxed{1}
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Worked example
A divergent improper integral
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1
Does \displaystyle\int_1^\infty \dfrac{1}{x}\,dx converge?
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2
\displaystyle\lim_{b\to\infty}\int_1^b \dfrac{1}{x}\,dx = \lim_{b\to\infty}\big[\ln x\big]_1^b = \lim_{b\to\infty}\ln b.
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3
\ln b\to\infty as b\to\infty — no finite limit exists. \boxed{\text{diverges}}
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Where this goes next
The Laplace transform, ahead
Every entry in a Laplace transform table is the result of evaluating one of these improper integrals once, for a specific f(t), so the reader never has to redo it by hand again. The transform's very existence, for a given s, is exactly this convergence question.
Check your understanding (counts toward your progress)
How is \displaystyle\int_0^\infty f(t)\,dt formally defined?
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This treats f itself as though it were its own antiderivative evaluated at endpoints, which isn't the definition.
An improper integral is a limit of areas, not an average value.
It's defined precisely as this limit -- it just may or may not converge.
Why
An improper integral with an infinite bound is defined as the limit of an ordinary definite integral as that bound is pushed to infinity.
A Laplace transform \int_0^\infty f(t)e^{-st}\,dt is being computed for f(t)=e^{3t}. For which values of s does this integral converge?
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For s\le3, e^{(3-s)t} doesn't decay, so the integral diverges.
This is backwards -- a smaller s makes the exponent 3-s larger, growing even faster.
The transform exists once s is large enough to outpace the function's own growth rate.
Why
The integrand becomes e^{(3-s)t}, which decays (and the integral converges) exactly when 3-s<0, i.e. s>3.
The integrand becomes e^{(3-s)t} -- find when this decays as t\to\infty.
Evaluate \displaystyle\int_0^\infty e^{-5t}\,dt.
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Why
\int_0^\infty e^{-kt}\,dt = 1/k for k>0; with k=5, the value is 1/5=0.2.
Correct answer
Also accepted:
Why
Use \int_0^\infty e^{-kt}\,dt=1/k for k>0.
Does \displaystyle\int_0^\infty e^{-2t}\,dt converge or diverge?
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Check the sign and factor again: \left[-\tfrac12e^{-2t}\right]_0^b\to 0-(-\tfrac12)=\tfrac12.
e^{-2t}\to0 as t\to\infty, so the integral settles to a finite value -- it converges.
This one has a closed-form antiderivative, so the limit can be evaluated exactly.
Why
\int_0^\infty e^{-2t}\,dt=\lim_{b\to\infty}\left[-\tfrac12e^{-2t}\right]_0^b = 0-(-\tfrac12)=\tfrac12.
Compute \lim_{b\to\infty}\left[-\tfrac12 e^{-2t}\right]_0^b.