3
$\begingroup$

If $z=\int_0^x \sin(x-s)f(s) \, \mathrm ds$ can anyone explain to me how you would compute $z'=\frac{\mathrm d}{\mathrm dx}\int_0^x \sin(x-s)f(s) \, \mathrm ds$

Or in general $z'=\frac{\mathrm d}{\mathrm dx}\int_0^x \! f(s,x) \, \mathrm ds$? Thanks!

I considering using Leibniz's rule, but I don't know if it can be applied to this situation.

$\endgroup$
  • 1
    $\begingroup$ $\sin(x-s)=\sin(x)\cos(s)-\sin(s)\cos(x)$ $\endgroup$ – Peđa Terzić Oct 25 '11 at 12:45
4
$\begingroup$

Looking at the wikipedia page of differentiation under the integral sign, you set $a(x) = 0$, $b(x) = x$ and $f(x, s) = \sin(x-s) f(s)$.

Alternatively, consider $z(x) = \int_0^x \sin(x-s) f(s) \mathrm{d} s$, and think of $z(x+ \delta x)-z(x)$:

$$ \begin{eqnarray} z(x+ \delta x)-z(x) &=& \int_0^{x+\delta x} \sin(x+\delta x-s) f(s) \mathrm{d} s - \int_0^x \sin(x-s) f(s) \mathrm{d} s \\ &=& \int_0^{x+\delta x} \sin(x+\delta x-s) f(s) \mathrm{d} s - \int_0^x \sin(x+\delta x-s) f(s) \mathrm{d} s + \\&& \int_0^x \sin(x+\delta x-s) f(s) \mathrm{d} s - \int_0^x \sin(x-s) f(s) \mathrm{d} s \\ &=& \int_{x}^{x+\delta x} \sin(x+\delta x-s) f(s) \mathrm{d} s + \int_0^x \left( \sin(x+\delta x -s) - \sin(x-s) \right) f(s) \mathrm{d} s \\ &\sim& \delta x \left( \sin(\delta x) f(x) + \int_0^x \cos(x-s) f(s) \right) \end{eqnarray} $$ In the limit $\lim_{\delta x \to 0} \frac{z(x+\delta x)-z(x)}{\delta x}$ the first term becomes zero, and you are get the answer: $$ z^\prime(x) = \int_0^x \cos(x-s) f(s) \mathrm{d} s $$

$\endgroup$
0
$\begingroup$

use general case of Leibniz integral rule. http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign

$\endgroup$
0
$\begingroup$

A general approach to this type of problems might be this: think of $z$ as the composition of the two functions $f:\mathbb R ^2 \to \mathbb R$ and $g:\mathbb R \to \mathbb R ^2$:$$f(x,y)=\int _a ^y p(x,t)\text d t,\qquad g(x)=(x,x).$$ So $z(x)=(f\circ g)(x)$. Applying the chain rule (note that $g'=(1,1)$) you get: $$z'(x)=\partial _x f (x,x)+\partial _y(x,x),$$ so the answer is $$z'(x)=\int _a ^x \partial _x p(x,t)\text d t+p(x,x).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.