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I ran into the following problem and its solution:

The integration by parts formula $$ \int_{a}^{b}u\frac{dv}{dx}\,dx=uv\bigg|_{a}^{b}-\int_{a}^{b}v\frac{du}{dx}\,dx $$ is known to be valid for functions $u(x)$ and $v(x)$, which are continuous and have continuous first derivatives. However, we will assume that $u$, $v$, $du/dx$, and $dv/dx$ are continuous only for $a\leqslant x\leqslant c$ and $c\leqslant x \leqslant b$; we assume that all quantities may have a jump discontinuity at $x=c$.

(a) Derive an expression for $\int_{a}^{b}u\,dv/dx\,dx$ in terms of $\int_{a}^{b}v\,du/dx\,dx$. $$ \int_{a}^{b}u\frac{dv}{dx}\,dx=uv\bigg|_{a}^{b}+uv\bigg|_{c^+}^{c^-}-\int_{a}^{b}v\frac{du}{dx}\,dx. $$

Could anyone clarify to me how this was obtained?

Edit: Following the advice of Muphrid, I obtained the following:

$$\begin{align} \int_{a}^{c^-}u\frac{dv}{dx}\,dx+\int_{c^+}^{b}u\frac{dv}{dx}\,dx&=uv\bigg|_{a}^{c^-}-\int_{a}^{c^-}v\frac{du}{dx}\,dx+uv\bigg|_{c^+}^{b}-\int_{c^+}^{b}v\frac{du}{dx}\,dx,\\ \int_{a}^{b}u\frac{dv}{dx}\,dx&=\color{red}{uv\bigg|_{a}^{c^-}+uv\bigg|_{c^+}^{b}}-\int_{a}^{b}v\frac{du}{dx}\,dx. \end{align}$$

What is the rule for combining the terms in red?

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2 Answers 2

up vote 4 down vote accepted

In response to your work from Muphrid:

$$\begin{align} \color{red}{uv\bigg|_{a}^{c^-}+uv\bigg|_{c^+}^{b}} &= uv(c^-) - uv(a) + uv(b) - uv(c^+) \\ &= uv(c^-) - uv(c^+) + uv(b) - uv(a) \\ &= uv\bigg|_{c^+}^{c^-} +uv\bigg|^b_a \end{align}$$

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I am sorry; I changed the limits of integration after you saw my edit, but your derivation follows exactly the same way. Thank you! –  Josué Nov 2 '12 at 22:12
    
I see that now, and went ahead and fixed my answer. Best of luck. –  mixedmath Nov 2 '12 at 22:14

I suggest using the usual integration by parts formula on the separate intervals $[a,c)$ and $(c, b]$ and then putting the two results together.

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