Make a family, then differentiate the family, integrals from easy to advanced lesson

The integrand $(x - 1)/\ln x$ is positive on $(0, 1)$ , because its numerator and denominator are both negative there, and its values stay below $1$ . So the integral $\int_0^1 \frac{x - 1}{\ln x}\,dx$ is a number between $0$ and $1$ , yet none of the earlier techniques produces an antiderivative. This page places the integral in a family $F(a)$ with a parameter $a$ , differentiates with respect to $a$ so that the logarithm cancels, justifies that step with an integrable bound and fixes the constant from a known value. The answer is $\ln 2 \approx 0.6931$ .

Key terms

Parameter family of integrals

Definition A parameter family of integrals is a function of a parameter $a$ defined by an integral in another variable,

F(a) = \int_c^d f(x, a)\,dx,

for the values of $a$ in a set $P$ where the integral converges. The limits $c < d$ may be infinite. The integration variable $x$ is used up by the integral, so $F$ depends on $a$ alone. To evaluate a hard integral this way, choose the family so that the target is $F(a_1)$ for one value $a_1$ , the partial derivative $\partial f/\partial a$ is easier to integrate than $f$ , and $F(a_0)$ is known at some $a_0$ .

In plain words Instead of one curve there is a whole collection of curves, one for each value of $a$ , and $F(a)$ is the area under the curve with that value. The difficult integral is one member of the collection, and the hope is that the way the area changes with $a$ is simpler than the area itself.

Example For the target integral, replace $x$ in the numerator by $x^a$ and keep the $-1$ ,

F(a) = \int_0^1 \frac{x^a - 1}{\ln x}\,dx,

for $a \ge 0$ . Then $F(1)$ is the target integral, and $F(0) = 0$ because $x^0 - 1 = 0$ for every $x$ in $(0, 1)$ . The power $x^a$ is the right choice because its derivative in $a$ is $x^a \ln x$ , which contains the logarithm of the denominator.

Removable endpoint limits

Definition Let $f$ be continuous on an open interval $(c, d)$ with finite endpoints. If the one-sided limits $\lim_{x \to c^+} f(x)$ and $\lim_{x \to d^-} f(x)$ exist as finite numbers, defining $f$ at $c$ and at $d$ by these limits gives a continuous function on $[c, d]$ . Its integral over $[c, d]$ is an ordinary definite integral, it equals the improper integral of $f$ over $(c, d)$ , and the values at the two endpoints do not affect it.

In plain words A formula that looks undefined at an endpoint, for example of the form $0/0$ , may still have a sensible value there. When both endpoint limits are finite numbers, the gaps can be filled, the graph becomes a continuous curve on a closed interval, and the integral needs no special treatment.

Example For the family integrand $(x^a - 1)/\ln x$ with $a \ge 0$ , put $t = \ln x$ , which tends to $0$ from below as $x \to 1^-$ . Then

\begin{aligned} &\lim_{x \to 1^-} \frac{x^a - 1}{\ln x} \\[0.4em] &= \lim_{t \to 0^-} \frac{e^{at} - 1}{t} = a, \end{aligned}

the derivative of $e^{at}$ at $t = 0$ . As $x \to 0^+$ with $a > 0$ , the numerator tends to $-1$ while $\ln x \to -\infty$ , so the quotient tends to $0$ , and for $a = 0$ the integrand is $0$ everywhere. For $a = 1$ the target integrand extends to a continuous function on $[0, 1]$ that increases from $0$ to $1$ , so its integral lies between $0$ and $1$ .

Differentiation under the integral sign

Definition Let $P$ be an interval of parameter values and $(c, d)$ an interval of $x$ values, bounded or not. Suppose that

the functions $f(x, a)$ and $\frac{\partial f}{\partial a}(x, a)$ are continuous for $c < x < d$ and $a$ in $P$ ,
the integral $F(a) = \int_c^d f(x, a)\,dx$ converges for every $a$ in $P$ ,
some function $g$ with $\int_c^d g(x)\,dx < \infty$ satisfies $\left\lvert \frac{\partial f}{\partial a}(x, a) \right\rvert \le g(x)$ for all $c < x < d$ and all $a$ in $P$ .

Then $F$ is differentiable on $P$ , with a one-sided derivative at an endpoint of $P$ that belongs to $P$ , and

F'(a) = \int_c^d \frac{\partial f}{\partial a}(x, a)\,dx.

When $[c, d]$ and $P$ are closed bounded intervals and $f$ and $\partial f/\partial a$ are continuous on the whole rectangle $[c, d] \times P$ , the bound is automatic, because a continuous function on a closed bounded rectangle is bounded and a constant is integrable over $[c, d]$ .

In plain words The rate at which an area changes with a parameter is the area under the rate at which the curve changes, provided one fixed function with finite area lies above all of those rates. The method is also known as Feynman’s trick.

Example For the family $F(a)$ with $a$ in $P = [0, \infty)$ , write $x^a = e^{a \ln x}$ for $0 < x < 1$ . Then

\frac{\partial}{\partial a}\, \frac{x^a - 1}{\ln x} = \frac{x^a \ln x}{\ln x} = x^a,

and $0 < x^a \le 1$ , so $g(x) = 1$ is an integrable bound on $(0, 1)$ . The theorem gives, for every $a \ge 0$ and with a right derivative at $a = 0$ ,

F'(a) = \int_0^1 x^a\,dx = \frac{1}{a + 1},

where the lower limit contributes nothing because $a + 1 > 0$ . The functions on $[0, \infty)$ with derivative $1/(a + 1)$ are $\ln(a + 1) + C$ , and $F(0) = 0$ gives $C = 0$ . So $F(a) = \ln(a + 1)$ for $a \ge 0$ , and the target integral is

\begin{aligned} &\int_0^1 \frac{x - 1}{\ln x}\,dx = F(1) \\[0.4em] &= \ln 2 \approx 0.6931. \end{aligned}

Integrable bound

Definition For a family $f(x, a)$ with $x$ in $(c, d)$ and $a$ in $P$ , an integrable bound for the parameter derivative is a function $g$ that does not depend on $a$ , with $\int_c^d g(x)\,dx < \infty$ and

\left\lvert \frac{\partial f}{\partial a}(x, a) \right\rvert \le g(x)

for all $x$ in $(c, d)$ and all $a$ in $P$ . Differentiability is a local property, so it is enough that each value $a_0$ in $P$ has such a bound for the $a$ in $P$ within some fixed distance of $a_0$ .

In plain words The bound is one ceiling that lies above the rate of change for every parameter value at once, and the area under the ceiling is finite. Without such a ceiling, the derivative of the integral and the integral of the derivative can differ, or the second one may not exist.

Example In the target family the bound is $g(x) = 1$ on $(0, 1)$ , since $0 < x^a \le 1$ for all $a \ge 0$ . Without a bound the step can fail. For every $a > 0$ the Dirichlet integral

\int_0^{\infty} \frac{\sin(ax)}{x}\,dx = \frac{\pi}{2}

does not depend on $a$ , so its derivative is $0$ . Differentiating inside the integral instead gives $\int_0^{\infty} \cos(ax)\,dx$ , which diverges. No integrable bound exists here, since a bound would have to lie above $\lvert \cos(a_0 x) \rvert$ , whose integral over $(0, \infty)$ is infinite.

Leibniz integral rule

Definition Let $f$ and $\partial f/\partial a$ be continuous on a rectangle $[c, d] \times [\alpha, \beta]$ , and let $u(a)$ and $v(a)$ be differentiable on $[\alpha, \beta]$ with values in $[c, d]$ . Then for $a$ in $[\alpha, \beta]$ , with one-sided derivatives at $\alpha$ and $\beta$ ,

\begin{aligned} &\frac{d}{da} \int_{u(a)}^{v(a)} f(x, a)\,dx \\[0.4em] &= f(v(a), a)\, v'(a) \\[0.4em] &\quad - f(u(a), a)\, u'(a) \\[0.4em] &\quad + \int_{u(a)}^{v(a)} \frac{\partial f}{\partial a}(x, a)\,dx. \end{aligned}

With fixed limits $u$ and $v$ the two boundary terms are zero, which is differentiation under the integral sign on a closed bounded interval.

In plain words When the limits move with the parameter, the area changes for two reasons. The curve changes height, which the integral of $\partial f/\partial a$ measures, and the edges move, which adds or removes a thin strip at each end with area about the height there times the distance the edge moves. In the target family the limits $0$ and $1$ stay fixed, so no edge terms appear.

Example For $a > 0$ let

G(a) = \int_0^a \frac{dx}{a + x}.

Directly, $G(a) = \ln(2a) - \ln a = \ln 2$ , so $G'(a) = 0$ . The rule applies on every rectangle $[0, \beta] \times [\alpha, \beta]$ with $0 < \alpha < \beta$ , where $f(x, a) = 1/(a + x)$ and its partial derivative are continuous, and it finds the same derivative as a balance of two terms,

\begin{aligned} G'(a) &= \frac{1}{2a} - \int_0^a \frac{dx}{(a + x)^2} \\[0.4em] &= \frac{1}{2a} - \frac{1}{2a} = 0. \end{aligned}

The moving upper edge contributes $f(a, a) \cdot v'(a) = 1/(2a)$ , and the falling heights inside take away the same amount.

Constant from a known value

Definition Let $F$ be continuous on an interval $P$ with $F'(a) = h(a)$ at every interior point, and let $H$ be continuous on $P$ with $H' = h$ there as well. Then $F(a) = H(a) + C$ for a single constant $C$ , and one known value $F(a_0)$ at a point $a_0$ of $P$ gives $C = F(a_0) - H(a_0)$ . The derivative alone determines $F$ only up to this constant, and the known value has to be found independently of the formula being derived.

In plain words Knowing how fast a quantity changes tells you its shape but not its starting height. One value where the integral is easy, often because the integrand vanishes there, pins down the height.

Example In the target family $F(a) = \ln(a + 1) + C$ , and at $a = 0$ the integrand $(x^0 - 1)/\ln x$ is $0$ on $(0, 1)$ , so $F(0) = 0$ . Since $\ln 1 = 0$ , the constant is $C = 0$ . Dropping this step and keeping, say, $\ln(a + 1) + 5$ would give $F(1) = \ln 2 + 5$ , which is impossible because the integrand stays below $1$ .

Check yourself

Evaluate $\int_0^1 \frac{x^2 - x}{\ln x}\,dx$ with the family $F(a) = \ln(a + 1)$ found on this page.

Show the answer

Write the integrand as the difference of two members of the family,

\frac{x^2 - x}{\ln x} = \frac{x^2 - 1}{\ln x} - \frac{x - 1}{\ln x}.

Both integrals converge, so the integral is $F(2) - F(1) = \ln 3 - \ln 2 = \ln \tfrac32 \approx 0.4055$ .

For $b > 0$ let

G(b) = \int_0^{\infty} \frac{e^{-x} - e^{-bx}}{x}\,dx.

Find $G(b)$ by differentiating with respect to $b$ . Which integrable bound justifies the step, and which known value fixes the constant?

Show the answer

The integrand has the removable limit $b - 1$ as $x \to 0^+$ and decays exponentially as $x \to \infty$ , so $G(b)$ converges. Its derivative with respect to $b$ is $e^{-bx}$ . One bound for all $b > 0$ is impossible, because such a bound would be at least $1$ for every $x$ , which is not integrable over $(0, \infty)$ . Instead fix any $b_0 > 0$ . For $b \ge b_0$ ,

0 < e^{-bx} \le e^{-b_0 x},

and $\int_0^{\infty} e^{-b_0 x}\,dx = 1/b_0$ is finite. Since $b_0 > 0$ was arbitrary, for every $b > 0$

\begin{aligned} G'(b) &= \lim_{t \to \infty} \int_0^t e^{-bx}\,dx \\[0.4em] &= \frac{1}{b}, \end{aligned}

and $G(b) = \ln b + C$ . The known value is $G(1) = 0$ , because the integrand is $0$ when $b = 1$ . Hence $C = 0$ and $G(b) = \ln b$ , for example $G(2) = \ln 2 \approx 0.6931$ .

Let $H(b) = \int_0^{\infty} e^{-x^2} \cos(bx)\,dx$ for real $b$ . Use differentiation under the integral sign, integration by parts and the value $H(0) = \sqrt{\pi}/2$ to find $H(b)$ .

Show the answer

The derivative of the integrand with respect to $b$ is $-x e^{-x^2} \sin(bx)$ , and its absolute value is at most $x e^{-x^2}$ for every $b$ . That bound is integrable, since $\int_0^{\infty} x e^{-x^2}\,dx = \tfrac12$ , so $H'(b)$ is the integral of $-x e^{-x^2} \sin(bx)$ from $0$ to $\infty$ . Integrate by parts with $u = \sin(bx)$ and $v = \tfrac12 e^{-x^2}$ , so that $dv = -x e^{-x^2}\,dx$ and $du = b\cos(bx)\,dx$ . The boundary term $uv = \tfrac12 e^{-x^2} \sin(bx)$ is $0$ at $x = 0$ and tends to $0$ as $x \to \infty$ . What remains is $-\int_0^{\infty} v\,du$ , which is $-\tfrac{b}{2}$ times the integral defining $H(b)$ , so

H'(b) = -\frac{b}{2}\, H(b).

By the product rule, the derivative of $e^{b^2/4} H(b)$ is $e^{b^2/4}\left(H'(b) + \tfrac{b}{2} H(b)\right) = 0$ , so $e^{b^2/4} H(b)$ is constant. Its value at $b = 0$ is $H(0) = \sqrt{\pi}/2$ , the half-line Gaussian integral. Therefore

\begin{aligned} &\int_0^{\infty} e^{-x^2} \cos(bx)\,dx \\[0.4em] &= \frac{\sqrt{\pi}}{2}\, e^{-b^2/4}, \end{aligned}

and for example $H(2) = \sqrt{\pi}/(2e) \approx 0.3260$ .

Make a family, then differentiate the family

Key terms

Parameter family of integrals

Removable endpoint limits

Differentiation under the integral sign

Integrable bound

Leibniz integral rule

Constant from a known value

Check yourself

How this lesson connects

Builds on