Real Analysis: Integrals

Integration Techniques

$u$-Substitution

The chain rule for the composition of two differentiable functions $f$ and $g$ states that

$$\dfrac{d}{dx}(f \circ g) = f'(g(x))g'(x).$$

In eligible integrands, the $g(x)$ and $g'(x)$ terms in an integrand may be replaced by $u$ and $u'$ variables to convert the integral from a more complicated function of $x$ to a simpler function of $u,$ which should ideally be easier to integrate. After integration, the antiderivative in terms of $u$ may then be converted back to an antiderivative in terms of $x.$ In more complicated integrals, this process of substitution can happen multiple times, thus requiring multiple re-substitutions after integration.

This process is popularly known as $u$-substitution due to the substitution of complicated terms for simpler ones and the nearly unanimous preference for the letter $u.$

Integration by Parts

The product rule for the derivative of the product of two differentiable functions $f$ and $g$ states that

$$(fg)' = f'g + fg'.$$

By applying the fundamental theorem of calculus to this identity, we can produce the corresponding identity for integrals, known as integration by parts:

$$\int f(x) g'(x) \, dx = f(x)g(x) - \int g(x) f'(x) \, dx.$$

Improper Integration

Definite integrals are defined for closed intervals of the form $[a, b].$ However, these closed intervals can be extended to $\overline{\mathbb{R}}$ to take the forms of $[-\infty, b],$ $[a, \infty],$ and $[-\infty, \infty]$ to allow the calculation of areas under infinite stretches of the number line. The process for doing so involves taking a limit, and the result is called an improper integral, defined for each of the three cases as

$$\begin{array}{rcl}\displaystyle \int\limits_a^{\infty} f(x) \, dx & = & \displaystyle \lim\limits_{b \rightarrow \infty} \int\limits_a^b f(x) \, dx\\ \displaystyle \int\limits_{-\infty}^b f(x) \, dx & = & \displaystyle \lim\limits_{a \rightarrow -\infty}\int\limits_a^b f(x) \, dx\\ \displaystyle \int\limits_{-\infty}^{\infty} f(x) \, dx & = & \displaystyle \lim\limits_{a \rightarrow -\infty} \lim\limits_{b \rightarrow \infty} \int\limits_a^b f(x) \, dx \end{array} $$

Problems

  1. $u$-Substitution: Show that

    $$\int\limits_a^b f'(g(x))g'(x) \, dx = \int\limits_{g(a)}^{g(b)} f'(u) \, du.$$

    By the chain rule, we see that

    $$\dfrac{d}{dx}f(g(x)) = f'(g(x))g'(x).$$

    Therefore, by the fundamental theorem of calculus,

    $$\int\limits_a^b f'(g(x))g'(x) \, dx = f(g(b)) - f(g(a)).$$

    By the same theorem,

    $$\int\limits_{g(a)}^{g(b)} f'(u) \, du = f(g(b)) - f(g(a)).$$

    Therefore

    $$\int\limits_a^b f'(g(x))g'(x) \, dx = \int\limits_{g(a)}^{g(b)} f'(u) \, du.$$

    Show Answer

  2. Integration by Parts: Show that if $f$ and $g$ are differentiable on $[a, b],$ then

    $$\int\limits_a^b f(x)g'(x) \,dx = f(b)g(b) - f(a)g(a) - \int\limits_a^b f'(x)g(x) \, dx.$$

    Let $h(x) = f(x)g(x).$ Then by the product rule,

    $$h'(x) = f'(x)g(x) + f(x)g'(x).$$

    By the fundamental theorem of calculus,

    $$\int\limits_a^b h'(x) \, dx = h(b) - h(a).$$

    Expanding $h$ shows that

    $$\int\limits_a^b f'(x)g(x) + f(x)g'(x) \, dx = f(b)g(b) - f(a)g(a).$$

    By additivity, we see that

    $$\int\limits_a^b f'(x)g(x) \, dx + \int\limits_a^b f(x)g'(x) \, dx = f(b)g(b) - f(a)g(a).$$

    Finally, a little subtraction gives us the desired form of the result:

    $$\int\limits_a^b f(x)g'(x) \, dx = f(b)g(b) - f(a)g(a) - \int\limits_a^b f'(x)g(x) \, dx.$$

    Show Answer

  3. Integration by Parts: Show that if $f$ and $f$ are differentiable, then

    $$\int f(x) g'(x) \, dx = f(x)g(x) - \int f'(x)g(x) \, dx.$$

    Note: This proof is very similar to the prior one.

    By the chain rule, we see that

    $$\dfrac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x).$$

    By the fundamental theorem of calculus, it follows that

    $$f(x)g(x) = \int f'(x)g(x) + f(x)g'(x) \, dx.$$

    Rearranging gives us the desired result:

    $$\int f(x)g'(x) \, dx = f(x)g(x) - \int f'(x)g(x) \, dx.$$

    Show Answer