Show that $g(s)$ is equal to $H(s+1)\ldots$ where $H$ is the Heaviside function Let $f(s)=H(s+1)-H(s-1)$ where $H(s)$ is the Heaviside function. I'm told that $$H(s)=\begin{cases} 0 & \text{if } s < 0 \\ \frac{1}{2} & \text{if } s=0 \\ 1 & \text{if } s>0 \end{cases}$$ and that $f(s)$ is an even function.
Also let $$g(s)=f(-s)-2ae^{-as}\int_{-\infty}^s e^{as'}f(-s') \, ds'$$
I'm asked to show that $g(s)=H(s+1)(2e^{-a(s+1)}-1)-H(s-1)(2e^{-a(s-1)}-1)$
So far I have this:
\begin{align*} g(s) & = f(s) -2e^{-as}\int_{-\infty}^s e^{as}f(s') \, ds' \\ & = H(s+1)-H(s-1)-2e^{-as}\left( f(s)\frac{e^{as}}{a} -\int_{-\infty}^sf'(s')\frac{e^{as'}}{a}\right) \end{align*}
I'm not really sure what to do from this point though.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
\mrm{g}\pars{s} & =
\,\mrm{f}\pars{-s} -
2a\expo{-as}\int_{-\infty}^{s}\expo{as'}\,\mrm{f}\pars{-s'}\,\dd s' =
\,\mrm{f}\pars{-s} -
2a\expo{-as}\int_{-s}^{\infty}\expo{-as'}\,\mrm{f}\pars{s'}\,\dd s'
\\[1cm] & =
\,\mrm{f}\pars{-s}
\\[5mm] & - 2a\expo{-as}\braces{%
\left.-\,{1 \over a}\,{\expo{-as'}}\,\mrm{f}\pars{s'}
\right\vert_{\ s'\ =\ -s}^{\ s'\ \to\ \infty} +
{1 \over a}\int_{-s}^{\infty}\expo{-as'}\,\bracks{\delta\pars{s' + 1} - \delta\pars{s' - 1}}\,\dd s'}
\\[1cm] & =
\mrm{f}\pars{-s}  -2a\expo{-as}\bracks{{1 \over a}\,\expo{as}\,\mrm{f}\pars{-s} +
{1 \over a}\,\expo{a}\,\mrm{H}\pars{s - 1} -
{1 \over a}\,\expo{-a}\,\mrm{H}\pars{s + 1}}
\\[5mm] & =
\overbrace{\bracks{-\,\mrm{H}\pars{-s + 1} + \,\mrm{H}\pars{-s - 1}}}
^{\ds{-\,\mrm{f}\pars{-s}}}\ -\
2\expo{-a\pars{s - 1}}\,\mrm{H}\pars{s - 1} +
2\expo{-a\pars{s + 1}}\,\mrm{H}\pars{s + 1}
\\[5mm] & =
\bracks{\mrm{H}\pars{s - 1} - \,\mrm{H}\pars{s + 1}} -
2\expo{-a\pars{s - 1}}\,\mrm{H}\pars{s - 1} +
2\expo{-a\pars{s + 1}}\,\mrm{H}\pars{s + 1}
\\[5mm] & =
\bbox[15px,#ffe,border:1px dotted navy]{\ds{%
\mrm{H}\pars{s + 1}\bracks{2\expo{-a\pars{s + 1}} - 1} -
\mrm{H}\pars{s - 1}\bracks{2\expo{-a\pars{s - 1}} - 1}}}
\end{align}
A: Seems to me like we should begin by figuring out what $f$ is. We have
$$f(s)=\begin{cases} 0 & \text{if } s < -1 \\ \frac{1}{2} & \text{if } s = -1 \\
1 & \text{if } -1 < s < 1 \\ \tfrac{1}{2} & \text{if } s = 1 \\ 0 &\text{if } s > 1 \end{cases}$$
Set
$$I(s) := \int_{-\infty}^se^{ax}f(-x)dx = -\int_{-s}^{\infty}e^{-ax}f(x)dx.$$
After some computation we find that
$$I(s)=\begin{cases} 0 &\text{if } s \leq -1 \\ \frac{1}{a}\left(e^{as} - e^{-a}\right) & \text{if } -1 < s < 1 \\ \frac{1}{a}\left(e^{a} - e^{-a}\right) &\text{if } s \geq 1 \end{cases}$$
We can rewrite $I$ as
$$I(s) = \frac{e^{as}}{a}f(s) + \frac{e^a}{a}H(s-1) - \frac{e^{-a}}{a}H(s + 1),$$
thus
\begin{equation*}
\begin{aligned}
g(s) &= f(s) + 2ae^{-as}I(s) \\
&= H(s + 1) - H(s - 1) + 2e^{-a(s - 1)}H(s - 1) - 2e^{-a(s + 1)}H(s + 1) \\
&= H(s - 1)(2e^{-a(s - 1)} - 1) - H(s + 1)(2e^{-a(s + 1)} - 1).
\end{aligned}
\end{equation*}
This is $-1$ times what you claimed, so one of us must have a sign error somewhere.
A: We use a convolution to rewrite the integral part of $g$.
\begin{equation*}
 2a\int_{-\infty}^{s}e^{-a(s-s')}f(s')\, ds' =  2a\int_{-\infty}^{\infty}H(s-s')e^{-a(s-s')}f(s')\, ds' =2a\varphi\ast f(s)
\end{equation*}
where $\varphi(s) = H(s)e^{-as}$. To evaluate the convolution we use the two-sided Laplace transformation ${\mathcal{L}}$ defined by
\begin{equation*}
{\mathcal{L}}f(u) = \int_{-\infty}^{\infty}e^{-us'}f(s')\, ds'.
\end{equation*}
Then
\begin{equation*}
\begin{array}{lcl}
H(s)&\overset{\mathcal{L}}{\longmapsto}&\dfrac{1}{u}\\
e^{as}H(s)&\overset{\mathcal{L}}{\longmapsto}&\dfrac{1}{u-a}\\[2ex]
H(s-b)&\overset{\mathcal{L}}{\longmapsto}&\dfrac{1}{u}e^{-bu}\\[2ex]
e^{as}H(s-b)&\overset{\mathcal{L}}{\longmapsto}&\dfrac{1}{u-a}e^{-b(u-a)}.
\end{array}
\end{equation*}
Then we are prepared to transform the convolution.
\begin{gather*}
a\varphi\ast f(s) \overset{\mathcal{L}}{\longmapsto}\dfrac{a}{u+a}\dfrac{e^{u}-e^{-u}}{u} = \left(\dfrac{1}{u}-\dfrac{1}{u+a}\right)\left(e^{u}-e^{-u}\right) =\\[2ex] \dfrac{1}{u}e^{u}-\dfrac{1}{u}e^{-u} - \dfrac{1}{u+a}e^{u+a}e^{-a}+\dfrac{1}{u+a}e^{-(u+a)}e^{a}.
\end{gather*}
After applying the inverse Laplace transformation we have
\begin{equation*}
a\varphi\ast f(s) = H(s+1)-H(s-1) - H(s+1)e^{-a(s+1)}+H(s-1)e^{-a(s-1)}.
\end{equation*}
Finally we get
\begin{equation*}
g(s) = H(s+1)(2e^{-a(s+1)}-1)-H(s-1)(2e^{-a(s-1)}-1).
\end{equation*}
