What is the inverse Laplace of F(s)?

How it works

The inverse Laplace transform of F(s) is the function f(t), t ≥ 0, with ∫ from 0 to ∞ of e^(−st) f(t) dt = F(s). A computer algebra system (nerdamer, open source) does the algebra, in the background after you start typing.

1. Partial fractions. The algebra writes F as partial fractions, checked equal to F at 20 points (to 1 part in 10⁹).
2. Each fraction T (a top-level term, with its sign) is inverted by the first of these that works:
   • Directly: the algebra inverts T. Its answer g(t) is checked: a numeric integral of e^(−st) g(t) (from t = 10⁻¹² to 60/s) must equal T at s = 4, 6 and 8, to 1 part in 10⁵. If that fails, the shift rule is tried: T(s + 10), with its top and bottom multiplied out by the algebra, is inverted to h(t) (checked the same way), and g(t) = e^(10t) h(t), simplified by the algebra (checked at 20 points).
   • A derivative: when T/s can be inverted as above to R(t) with R(0) = 0, then g(t) = R′(t), found by the algebra (checked against a numeric difference quotient at 20 points), because the transform of R′ is s times that of R, less R(0).
   • A pole at a: when T = c/(s − a)ⁿ: n is the whole number nearest log10(|T(10⁵)/T(10⁶)|), a comes from |T(s)|^(−1/n), a straight line in s, read at s = 40 and 50 and rounded to a multiple of 1/12, and T(s)(s − a)ⁿ must have the same value at s = 20, 30 and 60 as at 90 (to 1 part in 10⁹). Then T(s + a) (multiplied out) is inverted, and g(t) is e^(at) times it. If none works, the page gives the first reason it found.
3. f(t) is the sum of the g's with their signs, written by the page in the syntax you type (t as the variable; e^(−t) may appear as 1/e^t).
4. The whole f(t) is checked once more. Its own transform, ∫ f(t) e^(−st) dt from 0 to ∞, is worked out with the trapezoid rule in u = ln t (steps of 1/32, from t = e^−40 to 750/s), at s = σ + 20 and σ + 30, where σ is the largest real pole a found in step 2 (0 when none is larger). It must equal F(s) to 1 part in 10⁶. The check in step 2 at s = 4, 6 and 8 allows an error of 10⁻⁵ even where T is far smaller than that (after the shift rule T(s + 10) is tiny), so it can pass a wrong g; this one cannot. 1/(s² + 1)³ gets no answer: the three ways give a wrong f(t), which this check refuses.

A result that holds a number the algebra could only give rounded (a whole number past 2⁵³; a fraction p/q, p and q being the numbers that multiply its top and its bottom, such as 288557167/(342919925e), when q after removing its factors 2 and 5 is over 1,000,000, or is over 1 while |p| times it is over 10¹²; or a decimal with more than 12 significant digits) is not shown.

What you can type

• A number has at most 15 digits in a row. A computer number keeps only about 16 digits, so a longer one (9007199254740993) would stand for a nearby number (9007199254740992), and the page asks for fewer digits instead. Write very large or very small numbers with a power of ten (1e-20).
• F(s) uses the variable s. Numbers can have decimals; + − * / and ^ for powers; brackets; a number or bracket next to s multiplies (3s, 2(s + 1), s (s^2 + 4)).
• Constants pi and e; functions of the integral calculator (such as sqrt). Transforms with e^(−as) (steps and delays) cannot be inverted here.

What gets no answer

• A transform the algebra cannot split into partial fractions, or a fraction none of the three ways inverts (e^(−2s)/s).
• A step that fails its check, finds no formula, or takes over 3 seconds.

Worked examples by hand

F(s) = (3s + 2)/(s² − 3s + 2) (Trench, section 8.2, Example 8.2.4). s² − 3s + 2 = (s − 1)(s − 2), and (3s + 2)/((s − 1)(s − 2)) = −5/(s − 1) + 8/(s − 2). Inverting each: f(t) = 8e^(2t) − 5eᵗ, written 8 e^(2t) - 5 e^t.

F(s) = 1/(s² − 1) (Trench, section 8.2, Example 8.2.1(a)). 1/((s − 1)(s + 1)) = (1/2)/(s − 1) − (1/2)/(s + 1), so f(t) = eᵗ/2 − e^(−t)/2 = sinh t, written e^t/2 - 1/(2 e^t).

F(s) = (s² − 5s + 7)/(s + 2)³ (Trench, section 8.2, Example 8.2.7). Partial fractions give 1/(s + 2) − 9/(s + 2)² + 21/(s + 2)³, whose inverses are e^(−2t), −9t e^(−2t) and 21t² e^(−2t)/2. So f(t) = e^(−2t)(1 − 9t + 21t²/2), written 1/e^(2t) - 9t/e^(2t) + 21t^2/(2 e^(2t)).