Single-Variable-Calculus · Unit 17 · Video 2 · Interactive Practice

Functions With No Formula: erf, Fresnel and Li(x)

IKey Formulas

Statement	Formula	Holds when
The bell-curve integral	$F(x) = \int_0^x e^{-t^2}\,dt$ : $\ F(0) = 0$ , $\ F'(x) = e^{-x^2}$ , $\ F''(x) = -2x\,e^{-x^2}$	every real $x$ (FTC 2; $e^{-t^2}$ is continuous)
Even integrand from $0$ gives an odd function	$G(x) = \int_0^x f(t)\,dt \;\Longrightarrow\; G(-x) = -G(x)$	$f$ continuous and even, lower limit exactly $0$
End levels and erf	$\lim_{x\to\pm\infty} F(x) = \pm\frac{\sqrt{\pi}}{2}$ , $\ \mathrm{erf}(x) = \frac{2}{\sqrt{\pi}}F(x) \to \pm 1$	the value $\frac{\sqrt{\pi}}{2}$ is stated here and proved in the later lecture on volumes
Fresnel and logarithmic integrals	$S'(x) = \sin(x^2)$ , $\ C'(x) = \cos(x^2)$ ; $\ \mathrm{Li}(x) = \int_2^x \frac{dt}{\ln t}$ , $\ \mathrm{Li}'(x) = \frac{1}{\ln x}$ , $\ \mathrm{Li}''(x) = -\frac{1}{x(\ln x)^2}$	$S$ , $C$ : every real $x$ , with $S(0) = C(0) = 0$ ; $\ \mathrm{Li}$ : $x > 1$ , with $\mathrm{Li}(2) = 0$

Key Insight: A function defined by an integral is sketched from its definition alone: derivative from FTC 2, value $0$ at the starting point, second derivative for concavity, symmetry, then the behaviour at the ends. The integral itself supplies values to any accuracy.

IIVisualization 1 — Sketching $F$ From Its Definition

$F(x)$ is the signed area under $e^{-t^2}$ from $0$ to $x$ ; the height at $x$ is the slope of $F$ .

Upper limit x 1.00 Curve

IIIVisualization 2 — Fresnel Integrals: Signed Area That Wiggles

$S(x) = \int_0^x \sin(t^2)\,dt$ and $C(x) = \int_0^x \cos(t^2)\,dt$ rise where the integrand is positive and fall where it is negative.

Function Upper limit x 2.00

IVVisualization 3 — $\mathrm{Li}(x)$ Tracks the Primes

The smooth, increasing, concave-down curve $\mathrm{Li}(x)$ runs close to the prime-counting staircase $\pi(x)$ .

Range x 100

💡 The prime number theorem (Hadamard and de la Vallée Poussin, 1896) says $\pi(x)/\mathrm{Li}(x) \to 1$ as $x \to \infty$ ; here $\pi(x)$ counts primes and is not $3.14\ldots$ The Riemann hypothesis, still unproved, is equivalent to a sharp bound on the gap $\pi(x) - \mathrm{Li}(x)$ .

VQuiz Questions

Problem 1 · Read the Derivatives Off FTC 2

Given: $G(x) = \displaystyle\int_0^x e^{-t^4}\,dt$ . Find $G'(1)$ and $G''(1)$ .

✅ Correct! FTC 2 gives

G'(x) = e^{-x^4}

, and the chain rule gives

G''(x) = -4x^3 e^{-x^4}

; at

x = 1

these are

e^{-1}

and

-4e^{-1}

❌ That differentiates the integrand.

G'

is the integrand itself evaluated at

x

e^{-x^4}

; your pair is

G''(1)

and

G'''(1)

, one derivative too far.

❌ The factor $-2x$ belongs to $e^{-x^2}$ . Here the exponent is

-x^4

, whose derivative is

-4x^3

, so

G''(x) = -4x^3e^{-x^4}

❌ The lower limit contributes no term.

e^{-1} - 1

is the integrand at

1

minus the integrand at

0

; FTC 2 gives just

G'(x) = e^{-x^4}

, since the lower limit

0

is a constant.

Show solution

The integrand $e^{-t^4}$ is continuous, so FTC 2 applies:

G'(x) = e^{-x^4}, \qquad G''(x) = \frac{d}{dx}e^{-x^4} = -4x^3 e^{-x^4}.

G'(1) = e^{-1} \approx 0.368, \qquad G''(1) = -4e^{-1} \approx -1.472.

The same reading gives the whole sketch: $G' > 0$ everywhere, so $G$ is increasing; $G'' < 0$ for $x > 0$ and $G'' > 0$ for $x < 0$ , so the origin is an inflection point where the slope is $G'(0) = 1$ .

Problem 2 · When Is the Function Odd?

Given: four functions defined by integrals, each for all real $x$ . Decide which is odd, then use the symmetry.

Which function is odd?

Given $G(1) \approx 0.845$ , what is $G(-1)$ ?

✅ Correct!

e^{-t^4}

is even and the integral starts at

0

, so

G

is odd and

G(-1) = -G(1) \approx -0.845

❌ The integrand $t\,e^{-t^4}$ is odd, so $P$ is even. Substituting

t = -u

gives

P(-x) = \int_0^x (-u)e^{-u^4}(-du) = P(x)

❌ The integrand is even, but the integral starts at $1$ . An odd function must be

0

0

, and

Q(0) = -\int_0^1 e^{-t^2}\,dt \approx -0.747

❌ $e^{-(t-1)^2}$ is the bell curve moved to peak at $t = 1$ , which is not even. Numerically

R(1.3) \approx 1.038

but

R(-1.3) \approx -0.138

❌ Area is positive, but this integral runs backward. From

0

-1

the direction is reversed, so the equal mirror-image area counts as negative.

❌ Nothing cancels here.

G(-1)

is one region, from

0

back to

-1

, whose area equals that of the region from

0

1

; its sign is negative, not zero. Zero is the integral of an odd function over

[-1, 1]

❌ $1.690$ is the area over the whole interval $[-1, 1]$ .

G(-1)

covers only

0

back to

-1

, and backward accumulation is negative.

Show solution

Odd test. For $G$ , substitute $t = -u$ , $dt = -du$ ; the limits $t: 0 \to -x$ become $u: 0 \to x$ :

G(-x) = \int_0^{-x} e^{-t^4}\,dt = \int_0^x e^{-u^4}(-du) = -G(x).

$P$ fails because its integrand is odd (it comes out even), $Q$ fails because $Q(0) \ne 0$ , and $R$ fails because $e^{-(t-1)^2}$ is not even.

Value. $G(-1) = -G(1) \approx -0.845$ .

Problem 3 · A Moving Upper Limit $x^2$

Given: $A(x) = \displaystyle\int_0^{x^2} e^{-t^2}\,dt = F(x^2)$ , where $F(x) = \int_0^x e^{-t^2}\,dt$ and $\lim_{x\to\infty} F(x) = \frac{\sqrt{\pi}}{2}$ . Sketch $A$ from its definition.

What is $A'(x)$ ?

What does $A$ do at $x = 0$ ?

What is $\lim_{x\to-\infty} A(x)$ ?

✅ Correct!

A'(x) = 2x\,e^{-x^4}

changes sign from

-

+

0

, so

A

has a minimum

A(0) = 0

;

A

is even, and both ends level off at

\frac{\sqrt{\pi}}{2}

❌ The upper limit is $x^2$ , not $x$ .

A(x) = F(x^2)

, so

A'(x) = F'(x^2)\cdot 2x

, and

F'(x^2) = e^{-(x^2)^2}

❌ The chain-rule factor is missing.

A(x) = F(x^2)

, so

A'(x) = F'(x^2)\cdot\frac{d}{dx}(x^2) = e^{-x^4}\cdot 2x

❌ The integrand must be evaluated at the upper limit $x^2$ .

F'(x^2) = e^{-(x^2)^2} = e^{-x^4}

, not

e^{-x^2}

❌ $A$ is not $F$ .

A'(x) = 2x\,e^{-x^4}

is negative for

x < 0

and positive for

x > 0

: the slope changes sign, which makes

x = 0

an extremum.

❌ Check the sign of $A'$ .

A' < 0

to the left of

0

and

A' > 0

to the right, so

A

falls into

0

and rises out of it: a minimum.

❌ The exponential is positive, but the factor $2x$ is not.

A'(0) = 2\cdot 0\cdot e^{0} = 0

, and

A'

changes sign there.

❌ That copies the oddness of $F$ .

A(-x) = F((-x)^2) = F(x^2) = A(x)

, so

A

is even and both ends approach the same level.

❌ The slope tends to $0$ , not the value.

A(x) = F(x^2)

and

x^2 \to \infty

x \to -\infty

❌ $A$ grows as $x \to -\infty$ , but it is bounded.

A(x) = F(x^2) < \frac{\sqrt{\pi}}{2}

for every

x

, since the total area under

e^{-t^2}

from

0

\infty

is finite.

Show solution

Derivative. $A(x) = F(x^2)$ with $F'(u) = e^{-u^2}$ , so by the chain rule

A'(x) = F'(x^2)\cdot 2x = 2x\,e^{-x^4}.

At $0$ . $A(0) = \int_0^0 e^{-t^2}\,dt = 0$ . Since $e^{-x^4} > 0$ , the sign of $A'$ is the sign of $x$ : $A$ decreases on $(-\infty, 0)$ and increases on $(0, \infty)$ , so $x = 0$ is a local (in fact global) minimum.

Ends. $A(-x) = F(x^2) = A(x)$ , so $A$ is even. As $x \to \pm\infty$ , $x^2 \to \infty$ , so

\lim_{x\to-\infty} A(x) = \lim_{u\to\infty} F(u) = \frac{\sqrt{\pi}}{2}.

Problem 4 · Rescaling the Bell Curve

Given: $\lim_{x\to\infty}\int_0^x e^{-t^2}\,dt = \frac{\sqrt{\pi}}{2}$ and the wider bell curve used in statistics, $N(x) = \displaystyle\int_0^x e^{-t^2/2}\,dt$ . Find $\lim_{x\to\infty} N(x)$ .

✅ Correct! With

t = \sqrt{2}\,u

N(x) = \sqrt{2}\,F\!\left(\frac{x}{\sqrt{2}}\right) \to \sqrt{2}\cdot\frac{\sqrt{\pi}}{2} = \sqrt{\frac{\pi}{2}}

❌ The bell curve changed, so the area changes.

e^{-t^2/2} \ge e^{-t^2}

for every

t

, with equality only at

t = 0

, so the total area must be larger than

\frac{\sqrt{\pi}}{2}

❌ The factor $\sqrt{2}$ went the wrong way. With

t = \sqrt{2}\,u

we get

dt = \sqrt{2}\,du

, which multiplies the integral by

\sqrt{2}

; and a wider curve cannot have a smaller area.

❌ $\sqrt{2\pi}$ is the area over the whole line.

N

starts at

0

, so its limit is half of that, by the evenness of

e^{-t^2/2}

Show solution

Substitute $t = \sqrt{2}\,u$ , so $dt = \sqrt{2}\,du$ , $t^2/2 = u^2$ , and the limits $t: 0 \to x$ become $u: 0 \to x/\sqrt{2}$ :

N(x) = \int_0^{x/\sqrt{2}} e^{-u^2}\,\sqrt{2}\,du = \sqrt{2}\,F\!\left(\frac{x}{\sqrt{2}}\right).

As $x \to \infty$ , also $x/\sqrt{2} \to \infty$ , so

\lim_{x\to\infty} N(x) = \sqrt{2}\cdot\frac{\sqrt{\pi}}{2} = \frac{\sqrt{2\pi}}{2} = \sqrt{\frac{\pi}{2}} \approx 1.253.

This is the same move as the erf rescaling: a constant factor in front of $F$ scales both end levels by that factor.

Solved: 0 / 4

Functions With No Formula: erf, Fresnel and Li(x)

IKey Formulas

IIVisualization 1 — Sketching FFF From Its Definition

IIIVisualization 2 — Fresnel Integrals: Signed Area That Wiggles

IVVisualization 3 — Li(x)\mathrm{Li}(x)Li(x) Tracks the Primes

VQuiz Questions

IIVisualization 1 — Sketching $F$ From Its Definition

IVVisualization 3 — $\mathrm{Li}(x)$ Tracks the Primes