What Is the Laplace Transform of $t e^{at}\sin(bt)$?

For the unilateral Laplace transform, the answer follows from a standard differentiation rule: multiplying a function by $t$ corresponds to taking the negative derivative of its transform with respect to $s$. Applying that rule gives a rational function with a squared quadratic denominator.

Assume $a$ and $b$ are real constants and first take $b\ne 0$. The transform is

$$\mathcal{L}\{t e^{at}\sin(bt)\}(s) =\frac{2b(s-a)}{\big((s-a)^2+b^2\big)^2}, \qquad \operatorname{Re}(s)>a.$$

The condition on $s$ is part of the answer: it is the region where the defining integral converges. The case $b=0$ is separate because the original function is then identically zero.

Start with the transform of the sine

The unilateral Laplace transform is defined by

$$\mathcal{L}\{f(t)\}(s)=\int_0^\infty e^{-st}f(t)\,dt.$$

A standard transform pair is

$$\mathcal{L}\{\sin(bt)\}(s)=\frac{b}{s^2+b^2}, \qquad \operatorname{Re}(s)>0.$$

Multiplication by $e^{at}$ shifts the transform variable: replace $s$ by $s-a$. Therefore, for $g(t)=e^{at}\sin(bt)$,

$$G(s)=\mathcal{L}\{g(t)\}(s) =\frac{b}{(s-a)^2+b^2}, \qquad \operatorname{Re}(s)>a.$$

Apply the multiplication-by-t rule

If $G(s)=\mathcal{L}\{g(t)\}(s)$, then

$$\mathcal{L}\{t g(t)\}(s)=-\frac{d}{ds}G(s).$$

This rule can be seen by differentiating the defining integral: differentiating $e^{-st}$ with respect to $s$ contributes a factor of $-t$. Here, the integral and its derivative converge in the same half-plane, so the differentiation is valid for $\operatorname{Re}(s)>a$.

Differentiate the shifted sine transform using the chain rule:

$$\begin{aligned} \mathcal{L}\{t e^{at}\sin(bt)\}(s) &=-\frac{d}{ds}\left(\frac{b}{(s-a)^2+b^2}\right)\\ &=-b\left(-\frac{2(s-a)}{\big((s-a)^2+b^2\big)^2}\right)\\ &=\frac{2b(s-a)}{\big((s-a)^2+b^2\big)^2}. \end{aligned}$$

The two minus signs cancel. That sign is an easy detail to lose when using the rule.

Worked example

Find the transform when $a=2$ and $b=3$. Substitute these values into the general expression:

$$\mathcal{L}\{t e^{2t}\sin(3t)\}(s) =\frac{2(3)(s-2)}{\big((s-2)^2+3^2\big)^2} =\frac{6(s-2)}{\big((s-2)^2+9\big)^2}, \qquad \operatorname{Re}(s)>2.$$

For example, evaluating the transform at $s=4$ gives

$$\frac{6(4-2)}{\big((4-2)^2+9\big)^2} =\frac{12}{169}.$$

The restriction $s>2$ in this real-valued example is essential. The exponential factor in the integrand is $e^{-(s-2)t}$, and the integral does not converge when $s$ is at or to the left of the boundary $s=2$.

Convergence and special cases

For real $a$ and nonzero real $b$, the absolute value of the integrand is bounded by $t e^{-(\operatorname{Re}(s)-a)t}$. This bound is integrable on $[0,\infty)$ when $\operatorname{Re}(s)>a$, which gives the usual region of convergence. At the boundary, the factor no longer decays, and the improper integral does not converge in the ordinary sense.

If $b=0$, then $\sin(bt)=0$ for every $t$, so the transform is simply zero wherever the Laplace integral is defined. Do not substitute $b=0$ into the rational expression at a point where its denominator also vanishes; simplify the original function first.

Common mistakes

• Forgetting the negative sign in the rule: the correct identity is $\mathcal{L}\{t g(t)\}=-G'(s)$, not $G'(s)$.
• Missing the factor of two: differentiating $(s-a)^2$ produces $2(s-a)$.
• Leaving the denominator unsquared: differentiating a reciprocal quadratic gives $\big((s-a)^2+b^2\big)^2$ in the denominator.
• Dropping the convergence condition: the shift by $e^{at}$ moves the region from $\operatorname{Re}(s)>0$ to $\operatorname{Re}(s)>a$.
• Confusing $b$ with $b^2$ in the numerator: the sine transform begins with $b$, and differentiation contributes the factor $2(s-a)$; hence the numerator is $2b(s-a)$.