Calculate $\int_0^t e^{-s} \cos (t-s)\ \mathrm{d}\Theta(s)$ Calculate
$$\int_0^t e^{-s} \cos (t-s)\ \mathrm{d}\Theta(s)$$
where $\Theta$ is the Heaviside step function with the distributional derivative $\dot \Theta=\delta$.
My attempt. Since $\dot \Theta=\delta$ I think that the integral is different from $0$ only for $s=0$ and then:
$$\int_0^t e^{-s} \cos (t-s)\ \mathrm{d}\Theta(s)=\cos t$$
Is it true?
 A: $\newcommand{\d}{\,\mathrm{d}}$If you mean to use Riemann-Stieltjes integration, then we can proceed as follows. Any valid partition of $[0,t]$ must include some $[0=x_0,x_1]$. By definition: $$\int_0^tf(x)\d g(x):=\lim\sum f(t_i)(g(x_{i+1})-g(x_i))$$I use $t_i$ to denote the 'tags' in $[x_i,x_{i+1}]$ which are arbitrary. Note that, with $g=\Theta$, $\Theta(x_{i+1})-\Theta(x_i)=1-1=0$ if $i\ge1$ and it equals $1-0=1$ if $i=0$. Thus the RHS is always, for any partition, $f(t_0)\cdot(1)=f(t_0)$. As the partitions become finer, we are left with: $$\int_0^tf(s)\d\Theta(s)=\lim_{x_1\to0^+}f(t_0),\,0\le t_0\le x_1$$Over arbitrary choice of tags $t_0$. Assuming $f$ is sufficiently well-behaved - for the Riemann-Stieltjes integral to even be well-defined - we have that: $$\int_0^tf(s)\d\Theta(s)=\lim_{s\to0^+}f(s)\in\Bbb R$$For any $t>0$ and "nice" function $f$. In your case, of course $f$ is "nice" since it is continuous, and indeed $\lim_{s\to0^+}f(s)=f(0)$.
Conclusion: if $t>0$, $f:[0,t]\to\Bbb R$ is continuous, and if $\Theta$ is defined through $\Theta(0)=0,\,\Theta((0,\infty))=\{1\}$, then: $$\int_0^tf(s)\d\Theta(s)=f(0)$$As a Riemann-Stieltjes integral. I don't know distribution theory, and no distribution theory was required here.
So yes, under the above interpretation of the integral, the answer is $f(0)=\cos t$.
