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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?

Community
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wjmolina
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3 Answers3

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In response to your work from Muphrid:

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

davidlowryduda
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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.

Muphrid
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Corrected based on comment.

Consider the example $$ \int_{0}^2 x \, H(x-1) \; dx, $$ where $H(x)$ is the unit step function. Modifying the limits to account for $H(x-1)$ gives $$ \int_{0}^2 x \, H(x-1) \; dx= \int_{1}^2 x \; dx = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} $$

Applying standard integration by parts with $v = \frac{x^2}{2}$ and $u = H(x-1)$ gives the correct answer $$ \int_{0}^2 x \, H(x-1) \; dx= \frac{x^2}{2} H(x-1) \bigg|_{0}^2 - \int_{0}^2 \frac{x^2}{2} \delta(x-1) \; dx = \frac{4}{2} - 0 - \frac{1}{2} = \frac{3}{2}. $$

Since $u$ has a jump at $c=1$, the modified result proposed in this question would not capture the $-\frac{1}{2}$ from the delta function and instead get the same value from $$ \frac{x^2}{2} H(x-1) \bigg|_{1^+}^{1^-} = -\frac{1}{2}. $$

user3919899
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