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I have the next equation: $$\int_{0}^{t}h(\tau)e^{-(t-\tau)}\mathrm{d}\tau=10e^{-t}\cos(4t) \tag{1}$$

Derivating both sides, I get: $$h(t)e^{-(t-t)}=h(t)=10[(-1)e^{-t}\cos(4t)+e^{-t}(-4)\sin(4t)] \tag{2}$$

However, if first simplify (1) and then derivating: $$\int_{0}^{t}h(\tau)e^{\tau}\mathrm{d}\tau=10\cos(4t)$$ $$h(t)=10[-4\sin(4t)]e^{-t} \tag{3}$$

Please clarify me what is the correct expression among (2) and (3) and why.

Brallan Aguilar
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4 Answers4

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When you write $$\int_{0}^{t}h(\tau)e^{-(t-\tau)}d\tau,$$ what is the value of $t$ in the term $e^{-(t-\tau)}$? Is it the same value of $t$ that is the upper bound of the integral? Does it vary in the same way when you differentiate with respect to $t$?

As I interpret this formula, I see the following identity (with which I see you agree, since the way you "simplify" the formula implies this): $$\int_{0}^{t}h(\tau)e^{-(t-\tau)}d\tau = e^{-t} \int_{0}^{t}h(\tau)e^{\tau}d\tau.$$ But \begin{eqnarray} \frac{d}{dt}\left( e^{-t} \int_{0}^{t}h(\tau)e^{\tau}d\tau \right) & = & e^{-t} \frac{d}{dt} \left(\int_{0}^{t}h(\tau)e^{\tau}d\tau \right) + \left(\frac{d}{dt} e^{-t}\right) \int_{0}^{t}h(\tau)e^{\tau}d\tau \\ & = & e^{-t} (h(t)e^t) + (- e^{-t}) \int_{0}^{t}h(\tau)e^{\tau}d\tau \\ & = & h(t) - 10e^{-t} \cos(4t) \end{eqnarray} which is not equal to $h(t)$ except when $\cos(4t) = 0$. And oh, look, that term of $- 10e^{-t} \cos(4t)$ that is missing from your supposed derivative of $\int_{0}^{t}h(\tau)e^{-(t-\tau)}d\tau$ is exactly what you need to reconcile the (corrected) equation $(2)$ with equation $(3)$.

(Note that @copper.hat essentially gave this solution already in a comment on the original question; I am merely spelling it out in more detail.)

David K
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  • I suppose it accounts for missing $\cos(4t)$ factor (+1) ... I was wondering why eliminating $e^{-t}$ affected it. Also note that $h(0)$ does not seem to be zero. – S L May 28 '14 at 02:37
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If you wish to differentiate you should check Differentiation under integral sign. The integral on left side is known as Convolution of $h(t)*e^{-t}$ and it's Laplace transform has very nice form. Try taking Laplace transform on both sides and solve for $h(t)$ using inverse Laplace transform.

Using Laplace transform, I found that the solution to be $10 (\delta (t)-4 \sin (4 t)+\cos (4 t))$. Note that $h(0)$ isn't zero, you can find that by plugging $t=0$ on both sides where you get $0$ on left but $10$ on right side. While using FTC, it is important to note that you are forgetting to put he lower bound. i.e. $-e^{0}h(0) = -h(0)$. $h(0)$ accounts for $\delta (t)$.

S L
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  • I checked my calculus book and remember the Fundamental Theorem of Calculus. I think that (3) had a correct resolution but I'm not sure. – Brallan Aguilar May 28 '14 at 01:58
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    The first one can't really be done via FTC Part 1 because the upper bound on the integral is also in the integrand. Let me attempt to explain it using differentiation under the integral sign, if not, I will show a solution using the Laplace transformation. – FundThmCalculus May 28 '14 at 02:01
  • I also solved it using Laplace and is interesting the $\delta(t)$ term that is not get using FTC – Brallan Aguilar May 28 '14 at 02:48
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    @BrallanAguilar $$\lim_{\epsilon \to 0} \int_0^{\epsilon} h(\tau) e^{-(t-\tau)}d\tau = 10 = 10\lim_{\epsilon\to0}\int_{0}^{\epsilon} \delta(t) dt$$ since $e^{-(t-\tau)}$ is almost equal to $1$ and $h(t)$ is essentailly $h(0)$, it is only we guess that there should be delta function involved. – S L May 28 '14 at 02:53
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Solution using Differentiation Under the Integral Sign:

Definition (from the source all knowledge, Wikipedia ;)): $$\frac{d}{dx}=\left(\int\limits_{a(x)}^{b(x)}{f(x,t)dt}\right)=$$ $$f(x,b(x))*b'(x)-f(x,a(x))a'(x)+\int\limits_{a(x)}^{b(x)}{\frac{\partial f(x,t)}{\partial x}dt}$$ If the function doesn't depend on $x$, then we have the simpler solution that is the Fundamental Theorem of Calculus as taught in calculus textbooks, because $\frac{\partial f(x,t)}{\partial x}=0$.

In our case, we have the following: $a(t)=0$, $b(t)=t$, $f(t,\tau)=h(\tau)*e^{\tau-t}$. So we are going to now substitute. $$f(t,t)*(1)-f(t,0)*(0)+\int\limits_0^t {\frac{\partial \left( h(\tau)*e^{\tau-t} \right)}{\partial t}d\tau}$$ The $h(\tau)$ factors out because it is not dependent on $t$ (assumed constant). $$f(t,t)+\int\limits_0^t {h(\tau) \frac{\partial \left(e^{\tau-t} \right)}{\partial t}d\tau}$$ Take the partial derivative. $$f(t,t)+\int\limits_0^t {h(\tau)*e^{\tau-t}*(-1)d\tau}$$ Evaluate $f(t,t)$ $$h(t)+\int\limits_0^t {h(\tau)*e^{\tau-t}*(-1)d\tau}$$ Full equality: $$h(t)-\int\limits_0^t {h(\tau)*e^{\tau-t}*d\tau}=-10e^{-t}\cos(4t)-40e^{-t}\sin(4t)$$ Taking the original definition: $$h(t)=-10e^{-t}\cos(4t)-40e^{-t}\sin(4t)+\int\limits_0^t {h(\tau)*e^{\tau-t}*d\tau}$$ Substitution: $$h(t)=-10e^{-t}\cos(4t)-40e^{-t}\sin(4t)+10e^{-t}\cos(4t)$$ Simplify: $$h(t)=-40e^{-t}\sin(4t)$$ Thus, it appears that your second solution is correct.

FundThmCalculus
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An alternative approach (although I generally prefer simpler more straightforward approaches as in David K's answer above):

If $F(t) = \int_0^t f(s,t)ds$, then under appropriate smoothness assumptions we have $F'(t) = f(t,t) + \int_0^t {\partial f(s,t) \over \partial t } ds$. This is straightforward to establish using a Taylor expansion.

copper.hat
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