Turning the Convolution into a Finite Integral
One-sided exponential functions are simple to integrate, but their convolution depends on the overlap of two translated half-lines. Finding that overlap first prevents errors with the support and the sign of the answer. The equal-rate case also needs separate treatment: the formula for unequal rates has a denominator that vanishes when the rates coincide, even though the convolution itself remains finite.
Throughout, the measure space is \((\mathbb{R},\mathcal{L},m)\), where \(m\) is Lebesgue measure. This space is complete and sigma-finite, but not finite. For \(a>0\), define the real-valued measurable function \(f_a:\mathbb{R}\to[0,\infty)\) by
It is integrable, since \(\int_{\mathbb{R}} f_a\,dm=\int_0^\infty e^{-ax}\,dx=1/a\). We use the convolution convention \((f*g)(x)=\int_{\mathbb{R}}f(x-y)g(y)\,dm(y)\) whenever this integral is defined. The earlier existence result for \(L^1\) convolutions guarantees an integrable convolution defined \(m\)-a.e.; here the defining integral can be evaluated for every \(x\), giving a particular representative defined everywhere.
The Explicit Formula
Proof. Fix \(x\in\mathbb{R}\). The integrand is nonzero only if both \(y\ge0\) and \(x-y\ge0\). If \(x<0\), these inequalities cannot both hold, so the integrand vanishes for every \(y\) and \(h_{a,b}(x)=0\). If \(x=0\), they hold only at \(y=0\); a singleton has Lebesgue measure zero, so the integral is again zero. If \(x>0\), the conditions are exactly \(0\le y\le x\). Thus the integral is finite and equals $$ \int_0^x e^{-a(x-y)}e^{-by}\,dy =e^{-ax}\int_0^x e^{(a-b)y}\,dy. $$ When \(a=b\), the integrand in the last integral is \(1\), so the value is \(e^{-ax}x\). When \(a\ne b\), integrating the exponential gives $$ e^{-ax}\frac{e^{(a-b)x}-1}{a-b} =\frac{e^{-bx}-e^{-ax}}{a-b} =\frac{e^{-ax}-e^{-bx}}{b-a}. $$ The displayed expression also gives zero at \(x=0\), consistent with the direct integral there. The cases cover every real \(x\), proving the formula and its finiteness everywhere. \(\square\)
The order of the rates matters in the denominator as well as the numerator: if \(b>a\), then \(e^{-ax}\ge e^{-bx}\) for \(x\ge0\), and \(b-a>0\). If \(b<a\), both differences change sign. In either case, the convolution is nonnegative, as is also guaranteed directly by its nonnegative integrand.
Worked Example: Rates 2 and 5
Compute \(f_2*f_5\) everywhere. Since \(2\ne5\), for \(x\ge0\) the theorem gives $$ (f_2*f_5)(x)=\frac{e^{-2x}-e^{-5x}}{5-2} =\frac{e^{-2x}-e^{-5x}}{3}. $$ For \(x<0\), the value is \(0\). At \(x=0\), the formula gives \((1-1)/3=0\).
Check at \(x=\ln 2\): the formula gives \((1/4-1/32)/3=7/96\), since \(e^{-2\ln2}=1/4\) and \(e^{-5\ln2}=1/32\). Evaluating the defining integral at the same point gives $$ e^{-2\ln2}\int_0^{\ln2}e^{-3y}\,dy =\frac14\cdot\frac{1-e^{-3\ln2}}{3} =\frac14\cdot\frac{1-1/8}{3} =\frac{7}{96}. $$ The direct calculation agrees, and the answer at the boundary \(x=0\) is zero as required.
Worked Example: Equal Rates 3 and 3
For \(x\ge0\), the defining integral is $$ (f_3*f_3)(x)=\int_0^x e^{-3(x-y)}e^{-3y}\,dy =\int_0^x e^{-3x}\,dy =xe^{-3x}. $$ For \(x<0\), the integral is zero. In particular, at \(x=2\) the answer is \(2e^{-6}\).
Check: substituting \(x=2\) into the integral gives \(\int_0^2e^{-6}\,dy=2e^{-6}\), the same value. At \(x=0\), the interval of integration has length zero and the formula \(xe^{-3x}\) also gives zero. The equal-rate formula comes from integrating the constant \(e^{-3x}\) over an interval of length \(x\); it is not obtained by setting \(a=b\) in a quotient with denominator \(b-a\).
Worked Example: Rates 1 and 4
For \(x\ge0\), the unequal-rate formula gives $$ (f_1*f_4)(x)=\frac{e^{-x}-e^{-4x}}{4-1} =\frac{e^{-x}-e^{-4x}}{3}. $$ The value is zero for \(x<0\). To check at \(x=\ln2\), the formula gives \((1/2-1/16)/3=7/48\).
Directly, at \(x=\ln2\) the defining integral is $$ e^{-\ln2}\int_0^{\ln2}e^{-3y}\,dy =\frac12\cdot\frac{1-e^{-3\ln2}}{3} =\frac12\cdot\frac{1-1/8}{3} =\frac{7}{48}. $$ This agrees with the formula. For \(x>0\), the result is positive: the integral is of a strictly positive function over an interval of positive length. At \(x=0\), that interval has measure zero and the result is zero.
How the Two Cases Fit Together
Proof. Write \(h=b-a\), so \(h\to0\) and \(b>0\) for all sufficiently small \(|h|\). For \(h\ne0\), factor out \(e^{-ax}\) to obtain $$ \frac{e^{-ax}-e^{-(a+h)x}}{h} =e^{-ax}\frac{1-e^{-hx}}{h}. $$ If \(x=0\), this expression is zero for every \(h\ne0\), and its limit is \(0=xe^{-ax}\). If \(x>0\), set \(u=hx\). Then $$ e^{-ax}\frac{1-e^{-hx}}{h} =xe^{-ax}\frac{1-e^{-u}}{u}. $$ The elementary derivative of \(e^{-u}\) at \(u=0\) gives \(\lim_{u\to0}(1-e^{-u})/u=1\). Therefore the limit is \(xe^{-ax}\), as claimed. This is a limit of an explicit algebraic expression; no interchange of a limit and an integral is being used. \(\square\)
There is also a direct positivity check for the closed form. If \(x>0\) and \(a\ne b\), the numerator \(e^{-ax}-e^{-bx}\) has the same sign as \(b-a\), because the function \(r\mapsto e^{-rx}\) decreases as \(r\) increases. Their quotient is therefore positive. When \(a=b\), \(xe^{-ax}>0\) for \(x>0\). Together with the zero values for \(x\le0\), this matches the sign of the defining integral at every point.
Common Sign and Support Errors
A frequent error is to integrate the exponential and reverse only one difference. For \(a=2\), \(b=5\), an incorrect calculation might produce \((e^{-5x}-e^{-2x})/3\) for \(x>0\). At \(x=\ln2\), that is \((1/32-1/4)/3=-7/96\), which cannot be right because the integrand in the convolution is nonnegative. The correction is to retain the sign from the antiderivative: $$ e^{-2x}\frac{e^{-3x}-1}{-3} =\frac{e^{-2x}-e^{-5x}}{3}. $$ The corrected value at \(x=\ln2\) is \(7/96\), consistent with the direct integral.
A second error is to use the \(x\ge0\) formula on the whole real line. For example, extending the rates-2-and-5 expression to \(x=-1\) gives \((e^2-e^5)/3\), which is nonzero. But the defining integrand is zero for every \(y\) at \(x=-1\), since \(y\ge0\) and \(-1-y\ge0\) cannot both hold. The convolution is zero there. The support restriction is part of the formula, not an optional addition.
These computations illustrate why the convolution integral should be organized in two stages: first determine where both factors can be nonzero, and only then evaluate the exponential integral on that region. The equal-rate case is then transparent, while the limiting proposition explains why the two formulas join continuously as the rates coincide.
Check Your Understanding
Use the convention \((f*g)(x)=\int_{\mathbb{R}}f(x-y)g(y)\,dm(y)\), and assume all rates below are positive.
- For rates \(a=2\) and \(b=4\), determine the convolution for every real \(x\), and verify its value at \(x=\ln2\) directly from the integral.
- For \(a=b=1\), compute the convolution at \(x=0\) and \(x=3\). Explain why the value at \(x=0\) is not obtained by treating a singleton as an interval of positive measure.
- In the unequal-rate formula, explain why the numerator and denominator have the same sign for \(x>0\), whether \(b>a\) or \(b<a\).
- For fixed \(a>0\) and \(x>0\), compute the limit of \(\dfrac{e^{-ax}-e^{-bx}}{b-a}\) as \(b\to a\), and identify the resulting equal-rate convolution value.
- Explain why the unequal-rate expression alone does not give the convolution for negative \(x\), even though its algebraic formula can still be evaluated there.