Chapter 31Definition of the Laplace Transform and Basic Formulas

Proof

Substituting directly into Definition 31.1, we obtain

$$ \Lap[c_1f(t)+c_2g(t)] = \int_0^\infty \bigl(c_1f(t)+c_2g(t)\bigr)e^{-st}\,dt $$

Since integration preserves sums (a property of definite integrals, already covered in Chapter 5, Section 5.2), we can split the integral whose integrand is a sum into a sum of two integrals. Furthermore, the constants $c_1,c_2$ do not depend on the integration variable $t$, so they can be taken outside the integrals.

$$ = c_1\int_0^\infty f(t)e^{-st}\,dt + c_2\int_0^\infty g(t)e^{-st}\,dt = c_1\Lap[f(t)]+c_2\Lap[g(t)] $$

∎

Proof

Substitute $f(t)=1$ directly into Definition 31.1.

$$ \Lap[1] = \int_0^\infty 1\cdot e^{-st}\,dt = \lim_{c\to\infty}\int_0^c e^{-st}\,dt = \lim_{c\to\infty}\left[-\frac1s e^{-st}\right]_0^c $$

(The indefinite integral $\int e^{-st}\,dt=-\frac1s e^{-st}$ can be verified by differentiating the right-hand side with respect to $t$: this gives $-\frac1s\cdot(-s)e^{-st}=e^{-st}$, which returns the integrand.)

$$ = \lim_{c\to\infty}\left(-\frac1s e^{-sc} - \left(-\frac1s e^{0}\right)\right) = \lim_{c\to\infty}\left(-\frac1s e^{-sc}\right) + \frac1s $$

If $s>0$, then $e^{-sc}\to0$ as $c\to\infty$, so the first term converges to $0$. Hence

$$ \Lap[1] = 0+\frac1s = \frac1s\qquad(s>0) $$

∎ (For $s\le0$, $e^{-sc}$ does not converge to $0$, so the improper integral itself does not exist—this is what the condition “$s>0$” means.)

Proof

Substituting $f(t)=t$ into Definition 31.1 gives $\displaystyle\Lap[t]=\int_0^\infty t\cdot e^{-st}\,dt$. Since this is (unlike the constant function) a product of $t$ and $e^{-st}$, we use integration by parts (covered in high school mathematics), $\int u'v\,dt=uv-\int uv'\,dt$. Let $u=-\frac1s e^{-st}$ (this is an antiderivative of $e^{-st}$) and $v=t$; that is, $u'=e^{-st}$ and $v'=1$. Let us first carry out the computation in the form of an indefinite integral:

$$ \int t\cdot e^{-st}\,dt = \int t\cdot\left(-\frac1s e^{-st}\right)'dt = t\cdot\left(-\frac1s e^{-st}\right) - \int 1\cdot\left(-\frac1s e^{-st}\right)dt $$

(In the integration by parts formula, we chose $t$ as the factor to be differentiated and $e^{-st}$ as the factor to be integrated, because differentiating $t$ gives $1$ and so simplifies the expression.) Computing the integral that remains on the right-hand side, we get

$$ = -\frac{t}{s}e^{-st} + \frac1s\int e^{-st}\,dt = -\frac{t}{s}e^{-st} - \frac1{s^2}e^{-st} $$

Returning this to the form of a definite integral (an improper integral), we get

$$ \Lap[t] = \lim_{c\to\infty}\left[-\frac{t}{s}e^{-st} - \frac1{s^2}e^{-st}\right]_0^c $$

The part obtained by substituting the upper limit $t=c$ converges to $0$ as $c\to\infty$ when $s>0$ (the speed at which $e^{-sc}$ approaches $0$ is such that, even with a factor $c$ in front, $c\cdot e^{-sc}\to0$, and this is guaranteed by the fact that exponential growth is faster than polynomial growth—for this, one can use the idea of Taylor expansion in Chapter 4, or simply confirm $\lim_{c\to\infty}c/e^{sc}=0$ by l'Hôpital's rule). Substituting the lower limit $t=0$ gives $-\frac{0}{s}e^{0}-\frac1{s^2}e^{0}=-\frac1{s^2}$. Therefore,

$$ \Lap[t] = \Bigl(0-0\Bigr) - \left(-\frac1{s^2}\right) = \frac1{s^2}\qquad(s>0) $$

∎

Proof (by a recurrence relation and mathematical induction)

We use integration by parts for $f(t)=t^n$ ($n\ge1$). Letting $u=-\frac1se^{-st}$ ($u'=e^{-st}$) and $v=t^n$ ($v'=nt^{n-1}$), we get

$$ \Lap[t^n] = \int_0^\infty t^n e^{-st}\,dt = \underbrace{\left[t^n\cdot\left(-\frac1se^{-st}\right)\right]_0^\infty}_{0} + \frac1s\int_0^\infty n\,t^{n-1}e^{-st}\,dt $$

(Why the boundary term is $0$: substituting $t=0$ gives $t^n=0$ (since $n\ge1$), so the term is $0$. As $t\to\infty$, $t^n$ grows only polynomially, whereas $e^{-st}$ ($s>0$) approaches $0$ exponentially, so the product satisfies $t^n e^{-st}\to0$—this is just the same reasoning as in the proof for $\Lap[t]$, extended to degree $n$.)

Taking the constant $n$ on the right-hand side outside the integral, we get

$$ \Lap[t^n] = \frac{n}{s}\int_0^\infty t^{n-1}e^{-st}\,dt = \frac{n}{s}\Lap[t^{n-1}] $$

This is a recurrence relation (a formula that expresses a term by using the term one before it; here we apply the idea familiar from sequences to the sequence of functions given by “the $n$th Laplace transform”). Lowering $n$ one step at a time in the same formula, to $n-1,n-2,\ldots$, we obtain the following relations in succession:

$$ \Lap[t^{n-1}] = \frac{n-1}{s}\Lap[t^{n-2}],\qquad \Lap[t^{n-2}]=\frac{n-2}{s}\Lap[t^{n-3}],\ \ldots,\ \Lap[t^{1}]=\frac{1}{s}\Lap[t^0] $$

Substituting all of these into the first equation, we get

$$ \Lap[t^n] = \frac{n}{s}\cdot\frac{n-1}{s}\cdot\frac{n-2}{s}\cdots\frac{2}{s}\cdot\frac1s\cdot\Lap[t^0] $$

Here $\Lap[t^0]=\Lap[1]=\dfrac1s$ (Formula 31.1), and in the fraction $\dfrac ns\cdot\dfrac{n-1}s\cdots\dfrac1s$ the numerator is $n\times(n-1)\times\cdots\times1=n!$ (the factorial of $n$; $n!$ is read “n factorial”), while the denominator contains $n$ factors of $s$, so it is $s^n$. Therefore, we obtain

$$ \Lap[t^n] = \frac{n!}{s^n}\cdot\frac1s = \frac{n!}{s^{n+1}}\qquad(s>0) $$

(Strictly speaking, what we have shown is that “if the recurrence relation holds, then the general term is this”; to make the logic complete, the argument can be rewritten in the form of mathematical induction—“the formula holds for $n=0$ ($0!/s^1=1/s=\Lap[1]$, which is exactly Formula 31.1), and if we assume that it holds for $n=k$, then it also holds for $n=k+1$ by the recurrence relation.” The content is exactly the same as the computation above.) ∎

Proof

We substitute $f(t)=\sin\omega t$ into Definition 31.1 and integrate by parts (taking $u=-\frac1se^{-st}$, $v=\sin\omega t$, $v'=\omega\cos\omega t$).

$$ \Lap[\sin\omega t] = \int_0^\infty \sin\omega t\cdot e^{-st}\,dt = \underbrace{\left[\sin\omega t\left(-\frac1se^{-st}\right)\right]_0^\infty}_{0} + \frac{\omega}{s}\int_0^\infty \cos\omega t\cdot e^{-st}\,dt $$

(Why the boundary term is $0$: at $t=0$, $\sin0=0$. As $t\to\infty$, $|\sin\omega t|\le1$ (bounded), whereas $e^{-st}\to0$ (since $s>0$), so the product converges to $0$.) Applying integration by parts once more to the integral $\int_0^\infty\cos\omega t\,e^{-st}\,dt$ that remains on the right-hand side (taking $u=-\frac1se^{-st}$, $v=\cos\omega t$, $v'=-\omega\sin\omega t$), we obtain

$$ \int_0^\infty \cos\omega t\cdot e^{-st}\,dt = \underbrace{\left[\cos\omega t\left(-\frac1se^{-st}\right)\right]_0^\infty}_{\frac1s} - \frac{\omega}{s}\int_0^\infty \sin\omega t\cdot e^{-st}\,dt $$

(The boundary term this time: as $t\to\infty$ it is again $0$. At $t=0$, $\cos0\left(-\frac1se^{0}\right)=-\frac1s$, so $[\,\cdot\,]_0^\infty=0-\left(-\frac1s\right)=\frac1s$.) That is,

$$ \int_0^\infty \cos\omega t\cdot e^{-st}\,dt = \frac1s - \frac{\omega}{s}\Lap[\sin\omega t] $$

This is the equation we have obtained. Substituting it into the first equation, we get

$$ \Lap[\sin\omega t] = \frac{\omega}{s}\left(\frac1s - \frac{\omega}{s}\Lap[\sin\omega t]\right) = \frac{\omega}{s^2} - \frac{\omega^2}{s^2}\Lap[\sin\omega t] $$

This is a self-referential equation, in which the very quantity we want, $\Lap[\sin\omega t]$, also appears on the right-hand side. It looks strange at first, but it is nothing other than the equation $X=\dfrac{\omega}{s^2}-\dfrac{\omega^2}{s^2}X$ for the unknown $X=\Lap[\sin\omega t]$. We need only move terms to one side and solve for $X$.

$$ \left(1+\frac{\omega^2}{s^2}\right)\Lap[\sin\omega t] = \frac{\omega}{s^2} \qquad\therefore\quad \Lap[\sin\omega t] = \frac{\dfrac{\omega}{s^2}}{1+\dfrac{\omega^2}{s^2}} = \frac{\omega}{s^2+\omega^2} $$

(In the last equality, we multiplied the numerator and denominator by $s^2$ to clear the fraction within the fraction.) ∎

Proof

In the course of the proof of Formula 31.4, we already obtained the equation

$$ \int_0^\infty \cos\omega t\cdot e^{-st}\,dt = \frac1s - \frac{\omega}{s}\Lap[\sin\omega t] $$

Its left-hand side is exactly $\Lap[\cos\omega t]$, so we need only substitute the result of Formula 31.4, $\Lap[\sin\omega t]=\dfrac{\omega}{s^2+\omega^2}$, into the right-hand side.

$$ \Lap[\cos\omega t] = \frac1s - \frac{\omega}{s}\cdot\frac{\omega}{s^2+\omega^2} = \frac1s\left(1-\frac{\omega^2}{s^2+\omega^2}\right) = \frac1s\cdot\frac{(s^2+\omega^2)-\omega^2}{s^2+\omega^2} = \frac1s\cdot\frac{s^2}{s^2+\omega^2} = \frac{s}{s^2+\omega^2} $$

∎ (Because we had found Formula 31.4 first, Formula 31.5 required only a substitution—this is the reason for the order of computation in the original notes, $\sin$ first and $\cos$ afterward.)

Proof

Substituting $f(t)=e^{\omega t}$ into Definition 31.1 and using the law of exponents $e^{\omega t}\cdot e^{-st}=e^{-(s-\omega)t}$, we obtain

$$ \Lap[e^{\omega t}] = \int_0^\infty e^{\omega t}\cdot e^{-st}\,dt = \int_0^\infty e^{-(s-\omega)t}\,dt = \lim_{c\to\infty}\left[-\frac1{s-\omega}e^{-(s-\omega)t}\right]_0^c $$

If $s-\omega>0$ (that is, $s>\omega$), then as $c\to\infty$ we have $e^{-(s-\omega)c}\to0$ (for the same reason as in the proof of Formula 31.1), so

$$ \Lap[e^{\omega t}] = 0-\left(-\frac1{s-\omega}\right) = \frac1{s-\omega}\qquad(s>\omega) $$

∎ (We already checked the case $\omega=-2$ concretely in Example 31.1.)

Proof

We apply linearity (Theorem 31.1) to the definition (Chapter 2, Definition 2.2) $\sinh\omega t=\dfrac{e^{\omega t}-e^{-\omega t}}2$ and use Formula 31.6 (for both $\omega$ and $-\omega$).

$$ \Lap[\sinh\omega t] = \Lap\!\left[\frac{e^{\omega t}-e^{-\omega t}}2\right] = \frac12\Bigl(\Lap[e^{\omega t}]-\Lap[e^{-\omega t}]\Bigr) = \frac12\left(\frac1{s-\omega}-\frac1{s+\omega}\right) $$

(We used the fact that replacing $\omega\to-\omega$ in Formula 31.6 gives $\Lap[e^{-\omega t}]=\dfrac1{s-(-\omega)}=\dfrac1{s+\omega}$. Also, $\Lap[e^{\omega t}]$ exists for $s>\omega$ and $\Lap[e^{-\omega t}]$ exists for $s>-\omega$, so both exist simultaneously in the common range $s>|\omega|$.) Combining the fractions over a common denominator, we get

$$ \frac12\left(\frac1{s-\omega}-\frac1{s+\omega}\right) = \frac12\cdot\frac{(s+\omega)-(s-\omega)}{(s-\omega)(s+\omega)} = \frac12\cdot\frac{2\omega}{s^2-\omega^2} = \frac{\omega}{s^2-\omega^2} $$

∎

Proof

Exactly as for Formula 31.7, we apply linearity and Formula 31.6 to $\cosh\omega t=\dfrac{e^{\omega t}+e^{-\omega t}}2$.

$$ \Lap[\cosh\omega t] = \frac12\Bigl(\Lap[e^{\omega t}]+\Lap[e^{-\omega t}]\Bigr) = \frac12\left(\frac1{s-\omega}+\frac1{s+\omega}\right) = \frac12\cdot\frac{(s+\omega)+(s-\omega)}{s^2-\omega^2} = \frac12\cdot\frac{2s}{s^2-\omega^2} = \frac{s}{s^2-\omega^2} $$

∎

Proof

When $f(t)$ satisfies $|f(t)|\le Me^{\alpha t}$, we show that $\Lap[f(t)]$ exists (that is, the improper integral converges). Using the fact that the absolute value of an integral is at most the integral of the absolute value of the integrand (a property extending the triangle inequality to integrals), we have

$$ |F(s)| = \bigl|\Lap[f(t)]\bigr| = \left|\int_0^\infty f(t)e^{-st}\,dt\right| \le \int_0^\infty |f(t)|\,e^{-st}\,dt $$

Here, note that when $s>0$ we have $e^{-st}>0$, so the absolute value sign can be taken outside. Using the assumption $|f(t)|\le Me^{\alpha t}$ as well, we obtain

$$ \le \int_0^\infty Me^{\alpha t}e^{-st}\,dt = M\int_0^\infty e^{-(s-\alpha)t}\,dt $$

This last integral has exactly the same form as the one in the proof of Formula 31.1 (an integral with exponent $-(s-\alpha)$). If $s-\alpha>0$ (that is, $s>\alpha$), then

$$ M\int_0^\infty e^{-(s-\alpha)t}\,dt = M\left[-\frac1{s-\alpha}e^{-(s-\alpha)t}\right]_0^\infty = M\cdot\frac1{s-\alpha} = \frac{M}{s-\alpha} $$

(Since $s-\alpha>0$, we have $e^{-(s-\alpha)t}\to0$ as $t\to\infty$, and $e^0=1$ at $t=0$. This is simply a rerun of the proof of Formula 31.1.) We have thus shown that $|F(s)|\le\dfrac{M}{s-\alpha}$ (a finite value). Since the integral $\int_0^\infty|f(t)|e^{-st}\,dt$ is itself bounded above by a finite value (and $f$ is piecewise continuous, so $|f(t)|e^{-st}$ is integrable), the original improper integral $\int_0^\infty f(t)e^{-st}\,dt$ also converges. Therefore, if $f(t)$ satisfies $|f(t)|\le Me^{\alpha t}$, then $\Lap[f(t)]$ exists for every $s$ with $s>\alpha$. ∎

Proof

Substituting $f(t)=\delta(t-a)$ into Definition 31.1 gives $\displaystyle\Lap[\delta(t-a)]=\int_0^\infty \delta(t-a)e^{-st}\,dt$. For $a\ge0$, on the interval $(-\infty,0)$ we have $t-a<0$, so $\delta(t-a)=0$ (because the “impact” of the delta function is located only at the single point $t=a$). Therefore, even if we extend the interval of integration from $[0,\infty)$ to $(-\infty,\infty)$, the value of the integrand does not change, and neither does the value of the integral.

$$ \int_0^\infty \delta(t-a)e^{-st}\,dt = \int_{-\infty}^{\infty} e^{-st}\,\delta(t-a)\,dt $$

Now let $f(t)=e^{-st}$ and apply the sifting property of Definition 31.3; this picks out the value at $t=a$, namely $f(a)=e^{-sa}$.

$$ = f(a) = e^{-s\cdot a} = e^{-as} $$

∎

Proof

Substituting into Definition 31.1 gives $\displaystyle\Lap[u(t-a)]=\int_0^\infty u(t-a)e^{-st}\,dt$. Since $u(t-a)$ is $0$ for $t\lt a$ and $1$ for $t\ge a$, we split the interval of integration at $a$.

$$ \Lap[u(t-a)] = \int_0^a \underbrace{0}_{u(t-a)=0}\cdot e^{-st}\,dt + \int_a^\infty \underbrace{1}_{u(t-a)=1}\cdot e^{-st}\,dt = 0 + \int_a^\infty e^{-st}\,dt $$

The remaining integral has the same form as in the proof of Formula 31.1 (only the lower limit of the interval of integration is $a$ instead of $0$), so

$$ \int_a^\infty e^{-st}\,dt = \left[-\frac1se^{-st}\right]_a^\infty = 0-\left(-\frac1se^{-sa}\right) = \frac{e^{-as}}{s}\qquad(s>0) $$

∎

Proof

Substituting into Definition 31.1 and splitting the interval of integration at $a$, as in the proof of Formula 31.10, we obtain

$$ \Lap[f(t-a)u(t-a)] = \int_0^\infty f(t-a)u(t-a)e^{-st}\,dt = \int_0^a f(t-a)\cdot0\cdot e^{-st}\,dt + \int_a^\infty f(t-a)\,e^{-st}\,dt $$

(For $t\lt a$ we have $u(t-a)=0$, so the first term is $0$.) In the remaining integral, we make the substitution $u=t-a$ ($du=dt$; $u=0$ when $t=a$, and $u\to\infty$ as $t\to\infty$).

$$ \int_a^\infty f(t-a)e^{-st}\,dt = \int_0^\infty f(u)e^{-s(u+a)}\,du $$

(The change of the interval of integration, $[a,\infty)\to[0,\infty)$, corresponds to translating the $t$-axis by $a$ and reading it as the $u$-axis—you can think of it as simply shifting the horizontal axis by $a$ and placing the origin anew at $t=a$.) If we split the exponent as $e^{-s(u+a)}=e^{-sa}\cdot e^{-su}$, then $e^{-sa}$ is a constant that does not involve $u$, so it can be taken outside the integral.

$$ \int_0^\infty f(u)e^{-s(u+a)}\,du = e^{-sa}\int_0^\infty f(u)e^{-su}\,du = e^{-as}\Lap[f(t)] = e^{-as}F(s) $$

(Changing the symbol of the variable of integration from $u$ to $t$ does not change the value of the integral, so $\int_0^\infty f(u)e^{-su}\,du=\Lap[f(t)]=F(s)$.) ∎

Proof

Since $f(t)$ is a periodic function of period $T$, $f(t)=f(t+nT)$ holds ($n$ an integer). We divide the interval of integration $[0,\infty)$ of Definition 31.1 into single periods, $[0,T),[T,2T),[2T,3T),\ldots$.

$$ \Lap[f(t)] = \int_0^\infty f(t)e^{-st}\,dt = \underbrace{\int_0^T f(t)e^{-st}\,dt}_{F_0(s)} + \int_T^{2T} f(t)e^{-st}\,dt + \int_{2T}^{3T} f(t)e^{-st}\,dt + \cdots $$

We pull each of the integrals from the second term onward back to the first period by the substitution $u=t-nT$ ($n=1,2,\ldots$, $du=dt$). For example, in the second term ($n=1$), $u=0$ when $t=T$ and $u=T$ when $t=2T$, and by also using the periodicity $f(t)=f(u+T)=f(u)$, we obtain

$$ \int_T^{2T} f(t)e^{-st}\,dt = \int_0^T f(u+T)\,e^{-s(u+T)}\,du = e^{-sT}\int_0^T f(u)e^{-su}\,du = e^{-sT}F_0(s) $$

(Note that, since we changed the variable of integration from $t$ to $u$, the differential also changes from $dt$ to $du$.) Similarly, if we pull the $n$th term (the interval $[nT,(n+1)T)$) back by $u=t-nT$, then using the periodicity $n$ times we obtain $e^{-nsT}F_0(s)$. Therefore,

$$ \Lap[f(t)] = F_0(s) + e^{-sT}F_0(s) + e^{-2sT}F_0(s) + \cdots = F_0(s)\bigl(1+e^{-sT}+e^{-2sT}+\cdots\bigr) $$

The expression in parentheses is an infinite geometric series with common ratio $r=e^{-sT}$. If $s>0$, then $0\lt e^{-sT}\lt1$ (that is, $|r|\lt1$), so we can use the formula for the sum of an infinite geometric series, already covered in high school mathematics, $1+r+r^2+\cdots=\dfrac1{1-r}$ ($|r|\lt1$), and we obtain

$$ \Lap[f(t)] = F_0(s)\cdot\frac1{1-e^{-sT}} = \frac{F_0(s)}{1-e^{-sT}} $$

∎

Proof

(In the original notes (p. 13) this result is written in just one line with no proof, so the following is a proof supplied in this book.) We substitute $f'(t)$ into Definition 31.1 and integrate by parts. In the proof of Formula 31.2 ($\Lap[t]$) we went in the direction of “integrating $e^{-st}$ and differentiating $t$,” but this time, conversely, we differentiate $e^{-st}$ and integrate $f'(t)$. Letting $u=e^{-st}$ ($u'=-se^{-st}$) and $v'=f'(t)$ ($v=f(t)$), the integration by parts formula $\int_0^\infty uv'\,dt=\bigl[uv\bigr]_0^\infty-\int_0^\infty u'v\,dt$ gives

$$ \Lap[f'(t)] = \int_0^\infty f'(t)e^{-st}\,dt = \Bigl[f(t)e^{-st}\Bigr]_0^\infty - \int_0^\infty f(t)\cdot(-s)e^{-st}\,dt $$

We evaluate the boundary term. At $t=0$ it is $f(0)e^{0}=f(0)$. As $t\to\infty$, since $f(t)$ is of exponential order $\alpha$ ($|f(t)|\le Me^{\alpha t}$, Definition 31.2), we have $\bigl|f(t)e^{-st}\bigr|\le Me^{-(s-\alpha)t}$, and if $s>\alpha$ this tends to $0$ (by the same reasoning as in the proof of Theorem 31.2). Therefore, the boundary term equals

$$ \Bigl[f(t)e^{-st}\Bigr]_0^\infty = 0 - f(0) = -f(0) $$

Next, noting that the sign of the integral remaining on the right-hand side becomes $-\times(-s)=+s$, we find

$$ -\int_0^\infty f(t)\cdot(-s)e^{-st}\,dt = s\int_0^\infty f(t)e^{-st}\,dt = sF(s) $$

Combining the above, we obtain

$$ \Lap[f'(t)] = -f(0) + sF(s) = sF(s)-f(0) $$

∎

Proof (proof supplied in this book—as an application of the derivative formula, Theorem 31.4)

Let $f(t)=\mathrm{erf}(\sqrt{at})$. Then $f(0)=\mathrm{erf}(0)=0$. Combining the chain rule ($\dfrac{d}{dt}\sqrt{at}=\dfrac{a}{2\sqrt{at}}$) with $\mathrm{erf}'(x)=\dfrac2{\sqrt\pi}e^{-x^2}$, which we confirmed in the box above, we get

$$ f'(t) = \frac{2}{\sqrt\pi}e^{-(\sqrt{at})^2}\cdot\frac{d}{dt}\sqrt{at} = \frac2{\sqrt\pi}e^{-at}\cdot\frac{a}{2\sqrt{at}} = \frac{a}{\sqrt\pi\sqrt{at}}e^{-at} = \sqrt{\frac{a}{\pi t}}\,e^{-at} $$

(For the last equality, we simplified using $\dfrac{a}{\sqrt{at}}=\dfrac{a}{\sqrt a\sqrt t}=\dfrac{\sqrt a}{\sqrt t}$.) Noting here that $f(0)=0$, we use Theorem 31.4 to obtain

$$ \Lap[f'(t)] = sF(s) - f(0) = sF(s) $$

On the other hand, if we take the constant outside the integral, the left-hand side is

$$ \Lap[f'(t)] = \sqrt{\frac a\pi}\,\Lap\!\left[t^{-1/2}e^{-at}\right] = \sqrt{\frac a\pi}\int_0^\infty t^{-1/2}e^{-(s+a)t}\,dt $$

This integral can be computed with the substitution $t=u^2$ ($u\ge0$, $dt=2u\,du$, and $u:0\to\infty$ as $t:0\to\infty$).

$$ \int_0^\infty t^{-1/2}e^{-(s+a)t}\,dt = \int_0^\infty u^{-1}e^{-(s+a)u^2}\cdot2u\,du = 2\int_0^\infty e^{-(s+a)u^2}\,du $$

(Since $t^{-1/2}=(u^2)^{-1/2}=u^{-1}$ and $dt=2u\,du$, we get $u^{-1}\cdot2u=2$, so the $u$ cancels.) Next, with the substitution $x=\sqrt{s+a}\,u$ ($dx=\sqrt{s+a}\,du$), the integral takes a form to which the result of Chapter 7, Section 7.5 (Theorem 7.1, the Gaussian integral $\int_{-\infty}^\infty e^{-x^2}dx=\sqrt\pi$; since $e^{-x^2}$ is an even function, the one-sided integral is half of this, $\sqrt\pi/2$) applies directly.

$$ 2\int_0^\infty e^{-(s+a)u^2}\,du = \frac{2}{\sqrt{s+a}}\int_0^\infty e^{-x^2}\,dx = \frac2{\sqrt{s+a}}\cdot\frac{\sqrt\pi}2 = \frac{\sqrt\pi}{\sqrt{s+a}} $$

Therefore,

$$ \Lap[f'(t)] = \sqrt{\frac a\pi}\cdot\frac{\sqrt\pi}{\sqrt{s+a}} = \frac{\sqrt a}{\sqrt{s+a}} $$

Since this equals $sF(s)$, dividing both sides by $s$ gives

$$ F(s) = \frac1s\cdot\frac{\sqrt a}{\sqrt{s+a}} = \frac{\sqrt a}{s\sqrt{s+a}} $$

∎ (verified with mpmath: for $a=2,s=3$, the value of this expression agrees to 30 decimal places with the value obtained by direct numerical integration of $\mathrm{erf}(\sqrt{2t})e^{-3t}$ from $t=0$ to $\infty$). Simply by combining a single derivative formula (Theorem 31.4) with the Gaussian integral of Chapter 7—both tools already established before this chapter—we were able to give a complete proof of a result for which the original notes contained no proof. This is an instance of this book's policy of “omitting none of the content of the source” being carried through even in places where a proof is missing.