A calculus of the absurd

12.2 Integration by parts

When differentiating a function \(f(x) = a(x) \cdot b(x)\), the product rule (as in Section 11.4) tells us that

\begin{equation} \frac {d}{dx} \left [ f(x) \right ] = \frac {d}{dx} \left [a(x)\right ] b(x) + a(x) \frac {d}{dx} \left [b(x)\right ] \end{equation}

We can also integrate this expression,

\begin{equation} \int \frac {d}{dx} \left [ f(x) \right ] dx = \int \left ( \frac {d}{dx} \left [a(x)\right ] b(x) + a(x) \frac {d}{dx} \left [b(x)\right ] \right ) dx \end{equation}

As integration is linear8787 That is, the integral of a sum is the same as the sum of the integrals., this is equivalent to

\begin{equation} \int \frac {d}{dx} \left [ f(x) \right ] dx = \int \frac {d}{dx} \left [a(x)\right ] b(x) dx + \int a(x) \frac {d}{dx} \left [b(x)\right ] dx \end{equation}

As (by definition) \(\int \frac {df}{dx} dx = f(x)\), it is the case that

\begin{equation} f(x) = a(x) \cdot b(x) = \int \frac {d}{dx} \left [a(x)\right ] b(x) dx + \int a(x) \frac {d}{dx} \left [b(x)\right ] dx \end{equation}

Why is this useful? First we can do a little addition/rearranging

Note that \(\iff \) means “if and only if” - i.e. that the two equations are equivalent.

\begin{align} & a(x) \cdot b(x) = \int \frac {d}{dx} \left [a(x)\right ] b(x) dx + \int a(x) \frac {d}{dx} \left [b(x)\right ] dx \\ &\iff \int \frac {d}{dx} \left [a(x)\right ] b(x) dx = a(x) \cdot b(x) - \int a(x) \frac {d}{dx} \left [b(x)\right ] dx \label {int by parts formula} \end{align}

This gives us (a somewhat awkward) way to integrate products of expressions!

  • Example 19 Find the value of

    \begin{equation} \int z \sin (z) dz \end{equation}

First, fear not the fact that this is an integral in \(z\). As with sums, the variable itself does not matter (we could rename \(z\) to anything else, and the integral would still be the same integral)! Note that we could solve this without using integration by parts; we could think about what would differentiate to give us \(z \sin (z)\), but this is no fun (it’s much nicer to have a systematic way to solve a problem than to rely on lucky guessing)!

We start by trying to work out what should be what in the identity in Equation 12.6. We are starting with the left-hand side, and want to set \(\frac {d}{dz}\left [a(z)\right ] = \sin (z)\) and \(b(z) = z\). Why? Because in the integral on the right-hand side, we calculate the integral of the integral of \(a(z)\) and the derivative of \(b(z)\) and what makes the problem “hard” is not that we don’t know how to find the integral of either \(\sin (z)\) or \(z\), it’s that we don’t know how to find the integral of the two multiplied together! As the derivative of \(z\) is just \(1\), this gives us a way to rewrite the integral of the product of \(z\) and \(\sin (z)\) in terms of only the integral of \(\sin (z)\).

Let us proceed by using the formula for integration by parts (let us agree to set \(x=z\) if you’re not convinced that the variables can be renamed without changing the structure of the expressions), that is

\begin{equation} \int z \sin (z) dz = \left (\int \sin (z) dz\right ) z - \int \left (\int \sin (z) dz\right ) dz \end{equation}

The \(\int \sin (z) dz\) appears, because if \(\frac {d}{dz}\left [a(z)\right ] = \sin (z)\), then \(a(z) = \int \sin (z) dz\). We can proceed by simply evaluating the integrals on the right-hand side (all of which we know how to integrate)!

\begin{align} \int z \sin (z) dz &= -\cos (z) z - \int \left (-\cos (z)\right ) dz + C \\ &= -\cos (z) z + \int \cos (z) dz + C\\ &= -\cos (z) z + \sin (z) + C \end{align}

That this is the correct integral can be verified by differentiating the expression. You might also ask, why does the constant of integration \(C\) remain unchanged after each step (doesn’t each integral add something to it). The answer is that it doesn’t - at each step we are effectively redefining \(C\) to be \(C_{\text {new}} = C_{\text {old}} + k\) (where \(k\) is the constant of the new integral) - if we were to create a new constant each time it would get messy, so we can just “fold” all the constants into one new super-constant!