2
$\begingroup$

How do I differentiate definite integrals with the variable of differentiation as the upper or lower limit of integration / inside the integral such as the one below? $$ \frac{d}{dx} \int_a^x{(x-t)^n f^{(n+1)}(t)}dt $$ I know that integrating regular integrals i.e.: $ \frac{d}{dx} \int_a^x{f(t)}dt =f(x) $ Here, $x$ is the upper bound but $x$ is also within $ (x-t)^n $ so what happens to that? $$ $$ Edit: is it possible to have a general solution to this? Fundamental theorem of calculus says $ \frac{d}{dx} \int_a^x{f(t)}dt =f(x) $ but what about $ \frac{d}{dx} \int_a^x{f(t, x)}dt $ ? $$ $$ Edit2: obviously $$ \frac{d}{dx} \int_0^x {(x + t)^2}dt = \frac{d}{dx} \left(\frac{(x+t)^3}{3}\big|^x_0\right) = \frac{d}{dx}\frac{7x^3}{3} = 7x^2$$ but $$ \frac{d}{dx} \int_0^x {(x + t)^2}dt \not= (x+t)^2\big|^x_0 = 3x^2$$

$\endgroup$
1
  • $\begingroup$ You should have the answer $$\Gamma(n+1)(f'(x) + p_{n-1} ( x-a). $$ where $p_{n-1} $ is a polynomial of degree $n-1$. $\endgroup$ – Mhenni Benghorbal Apr 5 '16 at 1:40
1
$\begingroup$

Use integration by parts $n$ times. I'll get you started: let $u=(x-t)^n$ and $dv=f^{(n+1)}(t)\,dt$. Then $du=-n(x-t)^{n-1}$ and $v=f^{(n)}(t)$. Thus, $$\frac{d}{dx}\int_0^x(x-t)^nf^{(n+1)}(t)\,dt=\frac{d}{dx}\left((x-t)^nf^{(n)}(t)\big|^x_a+n\int_0^x (x-t)^{n-1}f^{(n)}(t)\,dt\right)$$ $$=-n(x-a)^{n-1}f^{(n)}(a)+n\frac{d}{dx}\int_0^x (x-t)^{n-1}f^{(n)}(t)\,dt$$

now do this $n-1$ more times (Hint: induction is your friend.)

$\endgroup$
2
  • $\begingroup$ Is that correct? Using the identity $ \int {u}dv = uv - \int {v}du $ means that $ \int {(x-t)^nf^{(n+1)}(t)}dt = f^{(n)}(t)(x-t)^n+n\int {(x-t)^{n-1}f^{(n)}(t)dt} $ which is different from what you have. $\endgroup$ – CloudIcarus Apr 5 '16 at 1:35
  • $\begingroup$ Thanks for the correction, I fixed it. $\endgroup$ – Alex S Apr 5 '16 at 1:37

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.