2
$\begingroup$

I only knew 1 kind of integration by parts given by this: $$\int u\, \Bbb dv = uv - \int v \,\Bbb du\tag{1}$$ And i can understand where it came from.

And the definite integral form is: $$\int_a^b u\, \Bbb dv = uv\,\Bigg|_a^b - \int_a^b v \,\Bbb du\tag{2}$$

From Stegun's Book, i found another interesting form of integration by parts given by: $$\int uv \Bbb dx = \left(\int u\Bbb dx\right)v - \int\left(\int u \Bbb dx\right)\frac{\Bbb dv}{\Bbb dx} \Bbb dx\tag{3}$$.

Maybe i can derive it using the product rule of differentation:

$$\begin{align} D_x\left[(u(x)v(x))\right] &= u'(x)v(x) + u(x)v'(x) \\ \int D_x\left[(u(x)v(x))\right] \Bbb dx&= \int u'(x)v(x) + u(x)v'(x)\Bbb dx\\ uv &= \int v \Bbb du + \int u \Bbb dv \end{align}$$

But what now?

My question is, what's the "definite" integral form of Eq. $(3)$? And how to prove it? (I confuse with $\frac{\Bbb dv}{\Bbb dx}$) and which parts are evaluated from $a$ to $b$ for the definite integral form?

Thanks in advance!

$\endgroup$
1
$\begingroup$

This is the same rule, it just uses different terms. In the second formula, $u$ is now $v$, and $v$ has become $\int u\,dx$, or put otherwise, $dv$ has become $u\,dx$.

$\endgroup$
2
  • $\begingroup$ What about the definite integral form? Is it like $$\int_a^b uv \Bbb dx = \left(\int_a^b u\Bbb dx\right)v - \int_a^b\left(\int_a^b u \Bbb dx\right)\frac{\Bbb dv}{\Bbb dx} \Bbb dx$$. ? $\endgroup$
    – user516076
    Jan 22 at 2:47
  • $\begingroup$ Are $v$ and $v'$ evaluated from $a$ to $b$ or not? $\endgroup$
    – user516076
    Jan 22 at 2:48

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.