Laplace Transform: Indeterminate Form in Definite Integral Change of Variables Calculation

Question

I was trying to find the Laplace transform of $e^{3t}$:

$$\int^\infty_0 e^{3t}e^{-st} \ dt = \int_0^\infty e^{3t - st} \ dt = \lim_{x \to \infty}\int_0^x e^{3t - st} \ dt$$

So if we then attempt to solve this using change of variables instead of integration by parts, and we change the bounds of integration by making the substitution $t = 0$ and $t = x$, then we get $u = 0$ and $u = 3x - sx$ as the new bounds. But we then get

$$\lim_{x \to \infty} \dfrac{1}{3 - s} \int_0^{3x - sx} e^u \ du = \dfrac{1}{3 - s} [(\infty - \infty) - 1],$$

which is an indeterminate form.

The solution should be $\dfrac{1}{s - 3}$.

What's going on here? Have I made an error?

I would appreciate it if people could please take the time to clarify this.

Answer 1

Let $S=s-3$.

Change of variables $u=(s-3)t=St$ which implies

$$du=S dt \iff dt=\frac{1}{S}du$$

permits to write the integral under the form :

$$\int_{u=0}^{u=+\infty} e^{-u}\frac{1}{S}du$$

whence the result : $\frac{1}{S}$

Answer 2

In fact we have $$\int_0^x e^{3t - st} \ dt={1\over 3-s}e^{(3-s)t}\Bigg|_{0}^{\infty}={1\over 3-s}\Big(e^{(3-s)\infty}-1\Big)$$which is convergent only when $\Re(s-3)>0\iff \Re(s)>3$ i.e. $s$ falls within ROC (Region of Convergence). In that case the answer would be $1\over s-3$.

Answer 3

$$\begin{align} \int^\infty_0 e^{3t}e^{-st} \ dt &= \int_0^\infty e^{3t - st} \ dt \\ &= \lim_{c \to \infty} \dfrac{1}{3 - s} \int_0^{c(3 - s)} e^u \ du \\ &= \lim_{c \to \infty} \dfrac{1}{3 - s} \left[ e^{c(3 - s)} - 1 \right] \\ &= \lim_{c \to \infty} \dfrac{1}{3 - s} \left[ e^{c(3 - (x + iy))} - 1 \right] \\ &= \lim_{c \to \infty} \dfrac{1}{3 - s} \left[ e^{c((3 - x) - iy))} - 1 \right] \\ &= \lim_{c \to \infty} \dfrac{1}{3 - s} \left[ e^{c(3 - x)} e^{-ciy} - 1 \right] \\ &= \lim_{c \to \infty} \dfrac{1}{3 - s} \left[ \dfrac{e^{c(3 - x)}}{e^{ciy}} - 1 \right] \\ &= \lim_{c \to \infty} \dfrac{1}{3 - s} \left[ \dfrac{e^{3c}e^{-cx}}{e^{ciy}} - 1 \right] \\ &= \lim_{c \to \infty} \dfrac{1}{3 - s} \left[ \dfrac{e^{3c}}{e^{ciy}e^{cx}} - 1 \right], \end{align}$$

which only converges when $x > 3$, since we then get $\lim_{c \to \infty} \dfrac{1}{3 - s} \left[ \dfrac{e^{cx}}{e^{ciy}} - 1 \right]$, and $e^{ciy}$ (Euler's formula) is bounded.

So the final result is $\dfrac{1}{3 - s} \left[ 0 - 1 \right] = -\dfrac{1}{3 - s} = \dfrac{1}{s - 3}$, as required.