## Archive for October 22nd, 2008

### Integration by parts via iterated integrals

October 22, 2008

(Another installment of an occasional series, featuring obfuscated proofs of elementary results.  Previously: Using calculus to generate the quadratic formula.  )

When faced with a challenging integration problem, sometimes some ingenuity is called for.  What follows is a technique that I vividly remember teaching in the late 1980s in a multivariable calculus course, which probably means this can be found in the 5th or 6th edition of Thomas and Finney.  (I don’t have a copy of that book anymore, so I can’t readily check.)

We’ll start with an example:  find $\int_{1}^4 \ln (x) \; dx$.

If you don’t know the antiderivative of the natural logarithm function, what to do?  Well, if you know its derivative, you might replace $\ln(x)$ with $\int_1^x \frac1t \; dt$.  If we make that substitution, we arrive at the double integral $\int_1^4 \; \int_1^x \frac{1}{t} \; dt \; dx$.

Next, we interchange the order of integration.

Graphing the region in the x,t plane, we see that the double integral can be rewritten as:

$\int_1^4 \; \int_1^x \; \frac1t \; dt \; dx \; = \; \int_1^4 \int_t^4 \frac1t \; dx \; dt$

This works out well, as now the inner antiderivative is simple.  We get

$\int_1^4 \int_t^4 \frac{1}{t} \; dx \; dt \; = \; \int_1^4 \frac{4-t}{t} \; dt$

which simplifies to

$\int_1^4 \frac{4}{t} - 1 \; dt = ( 4 \ln(4) - 4 ) - (4 \ln(1) - 1) = 4 \ln(4) - 3$.

I remember doing several examples like this for my students (as they’d been assigned as homework exercises), and thought it rather odd that every example in the text was one that could just as easily have been done using integration by parts.

And indeed, one can use this interated integral argument in place of integration by parts for any definite integral:

Consider $\int_a^b f(x) g'(x) \; dx$.  If we replace $f(x)$ by $\int_{f^{-1}(0)}^x f'(t) \; dt$, we find that

$\int_a^b f(x) g'(x) \; dx = \int_a^b \int_{f^{-1}(0)}^x \; f'(t) g'(x) \; dt \; dx$

Interchanging the order of integration is a bit tricky:

In general we end up with the sum of two iterated integrals:

$\int_{f^{-1}(0)}^a \int_a^b f'(t) g'(x) \;dx\;dt + \int_a^b \int_t^b f'(t) g'(x) \; dx\; dt$

The first integral yields

$\int_{f^{-1}(0)}^a f'(t) (g(b)-g(a)) \; dt = f(a)(g(b)-g(a))$.

The second integral yields

$\int_a^b f'(t) (g(b) - g(t)) \; dt = g(b)(f(b)-f(a)) - \int_a^b f'(t) g(t) dt$.

Adding these two answers gives the final result:

$\int_a^b f(x) g'(x) dx = f(b)g(b) - f(a)g(a) - \int_a^b f'(t) g(t) dt$

which is the same as the integration by parts formula:

$\int_a^b f(x) g'(x) dx = \left. f(x)g(x) \right|_a^b - \int_a^b f'(x) g(x) dx$

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