Recipe

Compute a convolution directly from the integral

Use when: you need (f*g)(t) and don't yet have F(s) and G(s) both in a form you can simply multiply and invert

Goal: a closed-form expression for (f*g)(t) as an ordinary function of t

Taught in Convolution and the convolution theorem

1. 1

   Write (f*g)(t)=\int_0^t f(\tau)g(t-\tau)\,d\tau: substitute the given f in as a function of \tau, and the given g with t-\tau plugged in for its input — not \tau itself.

   Example

   For f(\tau)=1, g(\tau)=e^{-2\tau}: (1*e^{-2t})(t)=\int_0^t 1\cdot e^{-2(t-\tau)}\,d\tau.

   Why: convolution is defined as this specific flip-and-slide integral, not as the ordinary product f(t)g(t) — mixing the two up is the single most common error with this operation.

2. 2

   Expand the integrand and separate any factor that depends only on t (not \tau) from the part that genuinely varies with \tau; the t-only factor comes straight out in front of the integral sign.

   Example

   e^{-2(t-\tau)}=e^{-2t}e^{2\tau}; since e^{-2t} doesn't depend on \tau, \displaystyle\int_0^t e^{-2t}e^{2\tau}\,d\tau = e^{-2t}\int_0^t e^{2\tau}\,d\tau.

   Why: this turns an expression mixing two variables into an ordinary single-variable integral in \tau, with t appearing only as a constant multiplier (and, next step, as the upper limit).

3. 3

   Find the antiderivative with respect to \tau, then evaluate it at \tau=t and \tau=0 and subtract — the fundamental theorem of calculus, only with a variable t sitting in the upper bound instead of a fixed number.

   Example

   \displaystyle\int_0^t e^{2\tau}\,d\tau=\left[\tfrac12e^{2\tau}\right]_0^t=\tfrac12(e^{2t}-1). Multiplying back the factor pulled out in Step 2: e^{-2t}\cdot\tfrac12(e^{2t}-1)=\tfrac12(1-e^{-2t}).

   • the integrand left after Step 2 is a pure exponential in \tau (as here) — integrate directly with the exponential antiderivative rule.
   • the integrand is a product needing integration by parts (one factor is \tau or t-\tau itself, not just an exponential or constant) — integrate by parts before evaluating the bounds.

   ⚡ Shortcut

   If F(s) and G(s) are already known (or easy to look up), skip this whole integral: compute the ordinary product F(s)G(s) and invert that instead — the convolution theorem itself, and usually the faster path once both transforms are already in hand.

   Why: this is the fundamental theorem of calculus applied with a parameter sitting in the bound — the result is a genuine function of t, exactly the form (f*g)(t) is supposed to take.