Improper integrals — Type I (infinite limits), Type II (discontinuous integrand): Worked examples

The scenario matrix

Every improper integral you meet falls into one of these cells. The two "bad-point types" are the columns; the "behaviour flavour" is the rows.

Cell	Bad point	Flavour	Example that hits it
A	Type I ( $\infty$ )	power decay, converges ( $p > 1$ )	Ex 1
B	Type I ( $\infty$ )	knife-edge, diverges ( $p = 1$ )	Ex 2
C	Type I ( $\infty$ )	exponential decay, converges	Ex 3
D	Type I ( $- \infty$ AND $+ \infty$ )	both halves, split required	Ex 4
E	Type I	signed integrand — principal-value trap	Ex 5
F	Type II (left-end $0$ )	mild spike, converges ( $p < 1$ )	Ex 6
G	Type II (interior blow-up)	hidden singularity, diverges	Ex 7
H	Type I ( $\infty$ ), no antiderivative	comparison decides it	Ex 8
I	real-world word problem	probability / physics	Ex 9
J	exam twist	both an $\infty$ end AND a spike at $0$	Ex 10

Recall The two decision rules to keep in your pocket

At infinity you need fast decay ( $p > 1$ ); at zero you need a mild spike ( $p < 1$ ). At $\infty$ , which $p$ converges? ::: $p > 1$ At $0$ , which $p$ converges? ::: $p < 1$ What happens at $p = 1$ ? ::: diverges in BOTH cases

Figure — Improper integrals — Type I (infinite limits), Type II (discontinuous integrand)

Cell A — Type I, power decay, converges

Worked example Example 1 —

$\int_{1}^{\infty} \frac{d x}{x ^{3/2}}$ Forecast: Here $p = \frac{3}{2}$ . Is $\frac{3}{2} > 1$ ? Guess convergence or divergence before reading on.

Step 1. Replace the forbidden $\infty$ with a real number $t$ : $\int_{1}^{\infty} x^{- 3/2} d x = lim_{t \to \infty} \int_{1}^{t} x^{- 3/2} d x .$ Why this step? We are not allowed to feed $\infty$ into the Fundamental Theorem of Calculus — it only works on a finite, closed interval $[1, t]$ . So we integrate to a movable wall $t$ first.

Step 2. Power rule, $\int x^{- 3/2} d x = \frac{x ^{- 1/2}}{- 1/2} = - \frac{2}{x}$ : $\int_{1}^{t} x^{- 3/2} d x = [- \frac{2}{x}]_{1}^{t} = - \frac{2}{t} + 2.$ Why this step? Now the interval $[1, t]$ is finite and the integrand is continuous there, so FTC is legal.

Step 3. Slide the wall to infinity. As $t \to \infty$ , $\frac{2}{t} \to 0$ (see Limits at Infinity): $lim_{t \to \infty} (2 - \frac{2}{t}) = 2.$ Why this step? Taking the limit is the final move that turns the movable-wall answer into the true infinite-area answer — we watch the leftover term $2/ t$ shrink to nothing.

Converges to $2$ . Cell A confirmed: $p = \frac{3}{2} > 1$ .

Verify: By the $p$ -integral formula the answer should be $\frac{1}{p - 1} = \frac{1}{3/2 - 1} = \frac{1}{1/2} = 2$ . ✓ Matches.

Cell B — Type I, knife-edge, diverges

Worked example Example 2 —

$\int_{2}^{\infty} \frac{d x}{x}$ Forecast: $p = 1$ exactly — the lonely loser. Predict the verdict.

Step 1. Movable wall: $\int_{2}^{\infty} \frac{d x}{x} = t \to \infty lim \int_{2}^{t} \frac{d x}{x}$ . Why this step? Same forbidden-infinity reason as Example 1 — $\infty$ is not a number we can feed to FTC, so we integrate to a real wall $t$ .

Step 2. The power rule fails at $p = 1$ (it would divide by $1 - p = 0$ ). The correct antiderivative of $1/ x$ is $ln x$ : $\int_{2}^{t} \frac{d x}{x} = [ln x]_{2}^{t} = ln t - ln 2.$ Why this step? $ln x$ is the one function whose derivative is $1/ x$ — this is exactly the case the power rule can't touch.

Step 3. As $t \to \infty$ , $ln t \to \infty$ : $lim_{t \to \infty} (ln t - ln 2) = \infty.$ Why this step? We slide the wall away and watch the area total; since $ln t$ has no ceiling, the area never settles on a number.

Diverges. Cell B confirmed. The logarithm grows without bound — slowly, but forever.

Verify: At $t = e^{10}$ , the partial area is $ln (e^{10}) - ln 2 = 10 - ln 2 \approx 9.307$ , and it keeps climbing with $t$ — no finite ceiling. ✓ Diverges.

Cell C — Type I, exponential decay

Worked example Example 3 —

$\int_{0}^{\infty} x e^{- x} d x$ Forecast: Exponential decay beats any polynomial. Guess: does $x e^{- x}$ enclose finite area?

Step 1. $\int_{0}^{\infty} x e^{- x} d x = t \to \infty lim \int_{0}^{t} x e^{- x} d x .$ Why this step? Infinite upper limit → replace with $t$ so FTC can act on the finite interval $[0, t]$ .

Step 2. Integrate by parts with $u = x$ , $d v = e^{- x} d x$ , so $d u = d x$ , $v = - e^{- x}$ : $\int_{0}^{t} x e^{- x} d x = [- x e^{- x}]_{0}^{t} + \int_{0}^{t} e^{- x} d x = - t e^{- t} + [- e^{- x}]_{0}^{t} = - t e^{- t} - e^{- t} + 1.$ Why this step? We can't antidifferentiate $x e^{- x}$ in one shot; integration by parts peels off the $x$ factor and leaves a plain $e^{- x}$ we know how to integrate.

Step 3. As $t \to \infty$ : $e^{- t} \to 0$ , and $t e^{- t} \to 0$ too because the exponential crushes the linear factor (see Limits at Infinity): $lim_{t \to \infty} (1 - t e^{- t} - e^{- t}) = 1 - 0 - 0 = 1.$ Why this step? This is where exponential decay earns its reputation — it drags the growing $t$ down to zero, so the leftover terms vanish and only the constant $1$ survives.

Converges to $1$ . Cell C confirmed.

Verify: This is the Gamma Function value $Γ (2) = 1! = 1$ . ✓ Matches.

Cell D — Type I on BOTH ends, split required

Worked example Example 4 —

$\int_{- \infty}^{\infty} \frac{d x}{1 + x ^{2}}$ Forecast: Both ends run to infinity. The rule says split and check both halves. Guess the total.

Step 1. Split at the convenient point $c = 0$ : $\int_{- \infty}^{\infty} \frac{d x}{1 + x ^{2}} = \int_{- \infty}^{0} \frac{d x}{1 + x ^{2}} + \int_{0}^{\infty} \frac{d x}{1 + x ^{2}} .$ Why this step? The definition demands each half converge independently — a single symmetric limit would only give a principal value.

Step 2. The antiderivative of $\frac{1}{1 + x ^{2}}$ is $arctan x$ . Right half: $\int_{0}^{\infty} = lim_{t \to \infty} [arctan x]_{0}^{t} = lim_{t \to \infty} (arctan t - 0) = \frac{π}{2} .$ Why this step? $arctan$ answers "which angle has this tangent?", and as $t \to \infty$ that angle approaches $9 0^{\circ} = \frac{π}{2}$ radians.

Step 3. Now the left half. We do not assume symmetry blindly — we compute it directly with its own movable wall $s \to - \infty$ : $\int_{- \infty}^{0} = lim_{s \to - \infty} [arctan x]_{s}^{0} = 0 - (- \frac{π}{2}) = \frac{π}{2} .$ Why this step? The definition requires each half to be tested on its own limit; here $arctan s \to - \frac{π}{2}$ as $s \to - \infty$ , and subtracting that negative gives $+ \frac{π}{2}$ . (It happens to match the right half because $\frac{1}{1 + x ^{2}}$ is even, but we earned that by computing, not by hand-waving.)

Both halves converge, so total $= \frac{π}{2} + \frac{π}{2} = π$ .

Verify: $arctan (\infty) - arctan (- \infty) = \frac{π}{2} - (- \frac{π}{2}) = π \approx 3.14159$ . ✓ Converges to $π$ .

Cell E — signed integrand, the principal-value trap

Worked example Example 5 —

$\int_{- \infty}^{\infty} x d x$ Forecast: The graph is symmetric about the origin, so "obviously" the positive and negative areas cancel to $0$ — right? Watch out.

Step 1. Split: $\int_{- \infty}^{\infty} x d x = \int_{- \infty}^{0} x d x + \int_{0}^{\infty} x d x .$ Why this step? Never merge into one symmetric limit — that only gives the Cauchy principal value, not true convergence.

Step 2. Test the right half alone: $\int_{0}^{\infty} x d x = lim_{t \to \infty} [\frac{x ^{2}}{2}]_{0}^{t} = lim_{t \to \infty} \frac{t ^{2}}{2} = \infty.$ Why this step? If even one piece diverges, the whole integral diverges — no need to compute the other, so we check the simplest half first.

Step 3. So $\int_{- \infty}^{\infty} x d x$ diverges, even though the symmetric limit $lim_{t \to \infty} \int_{- t}^{t} x d x = 0$ . Why this step? Stating the diverging conclusion right after one bad half is exactly the rule "one divergence kills the whole" in action — the symmetric $0$ is a mirage.

The trap: $0$ is the principal value, not the true value. Cell E confirmed.

Verify: Right-half partial area at $t = 1000$ is $\frac{100 0 ^{2}}{2} = 500000$ and grows without bound. ✓ Diverges despite the tempting symmetric $0$ .

Cell F — Type II, mild spike at the left end

Worked example Example 6 —

$\int_{0}^{1} \frac{d x}{x ^{2/3}}$ Forecast: The integrand blows up at $x = 0$ . Here $p = \frac{2}{3} < 1$ — a mild spike. Guess: finite area, or not?

Step 1. The bad point is the left end $x = 0$ , so creep up from $t > 0$ : $\int_{0}^{1} x^{- 2/3} d x = lim_{t \to 0^{+}} \int_{t}^{1} x^{- 2/3} d x .$ Why this step? $1/ x^{2/3} \to \infty$ as $x \to 0^{+}$ ; FTC can't touch $x = 0$ , so we tiptoe to it from the right.

Step 2. Power rule, $\int x^{- 2/3} d x = \frac{x ^{1/3}}{1/3} = 3 x^{1/3}$ : $\int_{t}^{1} x^{- 2/3} d x = [3 x^{1/3}]_{t}^{1} = 3 - 3 t^{1/3} .$ Why this step? On $[t, 1]$ the integrand is continuous, so FTC applies legally.

Step 3. As $t \to 0^{+}$ , $t^{1/3} \to 0$ : $lim_{t \to 0^{+}} (3 - 3 t^{1/3}) = 3.$ Why this step? Sliding $t$ down to the spike, the correction term $3 t^{1/3}$ vanishes — the tall-but-skinny sliver holds only finite area.

Converges to $3$ . Cell F confirmed: $p = \frac{2}{3} < 1$ .

Verify: Type II $p$ -integral gives $\frac{1}{1 - p} = \frac{1}{1 - 2/3} = \frac{1}{1/3} = 3$ . ✓ Matches.

Cell G — Type II, hidden interior singularity

Worked example Example 7 —

$\int_{- 1}^{2} \frac{d x}{x ^{2}}$ Forecast: Tempting to just write $[- 1/ x]_{- 1}^{2}$ . But is $x^{2}$ well-behaved on all of $[- 1, 2]$ ? Look for a blow-up.

Step 1. Spot the singularity: $\frac{1}{x ^{2}} \to + \infty$ at $x = 0$ , which lies inside $[- 1, 2]$ . Why this step? Before reaching for any antiderivative, scan the interval for points where the integrand explodes. This one hides at the origin.

Step 2. Split at $c = 0$ and require both halves to converge: $\int_{- 1}^{2} \frac{d x}{x ^{2}} = \int_{- 1}^{0} \frac{d x}{x ^{2}} + \int_{0}^{2} \frac{d x}{x ^{2}} .$ Test the right half: with $p = 2 \geq 1$ at the endpoint $0$ , $\int_{0}^{2} x^{- 2} d x = lim_{t \to 0^{+}} [- \frac{1}{x}]_{t}^{2} = lim_{t \to 0^{+}} (- \frac{1}{2} + \frac{1}{t}) = + \infty.$ Why this step? An interior blow-up breaks FTC across the whole interval, so we split at it; then $\frac{1}{t} \to + \infty$ shows the spike $1/ x^{2}$ is too tall to fit finite area.

Step 3. One half diverges → the whole integral diverges. Cell G confirmed. Why this step? By the "one divergence kills the whole" rule, we can stop the moment a single piece fails.

The trap: the naive plug-in $[- 1/ x]_{- 1}^{2} = - \frac{1}{2} - 1 = - \frac{3}{2}$ gives a negative number for a strictly positive integrand — an impossible negative area, the red flag that FTC was applied illegally across a discontinuity.

Verify: Right-half partial at $t = 1 0^{- 6}$ is $- \frac{1}{2} + 1 0^{6} = 999999.5$ , exploding as $t \to 0$ . ✓ Diverges (and the naive $- \frac{3}{2}$ is nonsense).

Cell H — Type I, no antiderivative, use comparison

Worked example Example 8 — Does

$\int_{1}^{\infty} \frac{d x}{x ^{4} + 1}$ converge? Forecast: No elementary antiderivative, but the interval runs to infinity so it is Type I. Since $x^{4} + 1 \approx x^{2}$ for large $x$ , guess the verdict, then justify it cleanly.

Step 1. Bound the integrand by something we CAN integrate. For $x \geq 1$ , since $x^{4} + 1 > x^{4}$ : $x^{4} + 1 > x^{4} = x^{2} ⟹ \frac{1}{x ^{4} + 1} < \frac{1}{x ^{2}} .$ Why this step? The Comparison Test for Integrals says: if a bigger non-negative function has finite area, the smaller one does too — so we hunt for a simple upper bound.

Step 2. The dominating integral converges — it is the $p$ -integral with $p = 2 > 1$ (this is exactly the master $\int_{1}^{\infty} x^{- p} d x$ result from the parent note, and it also underlies the p-series and Integral Test): $\int_{1}^{\infty} \frac{d x}{x ^{2}} = lim_{t \to \infty} [- \frac{1}{x}]_{1}^{t} = lim_{t \to \infty} (1 - \frac{1}{t}) = 1 < \infty.$ Why this step? We need a known finite value to compare against; $1/ x^{2}$ is the closest simple curve that dominates ours and has $p > 1$ , so its area is finite.

Step 3. Since $0 < \frac{1}{x ^{4} + 1} < \frac{1}{x ^{2}}$ and the upper integral is finite, our integral converges (to some value $\leq 1$ ). Cell H confirmed. Why this step? A non-negative area trapped strictly under a finite area must itself be finite — that is the entire logic of the comparison test.

Verify: Our integral is squeezed below $\int_{1}^{\infty} x^{- 2} d x = 1$ and above $0$ , so it is finite. (Numerically it is $\approx 0.927$ , comfortably in $(0, 1)$ .) ✓ Converges.

Cell I — real-world word problem

Worked example Example 9 — Exponential lifetime (probability)

A machine's failure time $X$ (in years) has probability density $f (x) = 2 e^{- 2 x}$ for $x \geq 0$ . (a) Confirm this is a valid density (total probability $= 1$ ). (b) Find the mean lifetime $\int_{0}^{\infty} x f (x) d x$ . Forecast: For a valid density the area under $f$ over all $x$ must be exactly $1$ . Guess the mean before computing.

Step 1 (a). Total probability: $\int_{0}^{\infty} 2 e^{- 2 x} d x = lim_{t \to \infty} [- e^{- 2 x}]_{0}^{t} = lim_{t \to \infty} (1 - e^{- 2 t}) = 1.$ Why this step? A density must enclose total area $1$ ; the infinite upper limit makes this a Type I improper integral, so we use the movable-wall limit. Valid.

Step 2 (b). Mean $= \int_{0}^{\infty} x \cdot 2 e^{- 2 x} d x$ . Integrate by parts ( $u = x$ , $d v = 2 e^{- 2 x} d x$ , $v = - e^{- 2 x}$ ): $\int_{0}^{t} 2 x e^{- 2 x} d x = [- x e^{- 2 x}]_{0}^{t} + \int_{0}^{t} e^{- 2 x} d x = - t e^{- 2 t} + [- \frac{1}{2} e^{- 2 x}]_{0}^{t} .$ Why this step? The $x$ factor demands integration by parts, just like Cell C — it strips the $x$ off and leaves a clean exponential.

Step 3. As $t \to \infty$ : $t e^{- 2 t} \to 0$ and $e^{- 2 t} \to 0$ : $mean = 0 + (\frac{1}{2} - 0) = \frac{1}{2} .$ Why this step? Exponential decay again defeats the linear $t$ , so all the wall-dependent terms vanish and only $\frac{1}{2}$ remains.

Mean lifetime $= \frac{1}{2}$ year (units: years — the integrand is $x [yr] \cdot f [yr^{- 1}] \cdot d x [yr]$ , giving years). Cell I confirmed. This connects to the Laplace Transform, which is exactly $\int_{0}^{\infty} e^{- s x} f (x) d x$ .

Verify: For rate $λ = 2$ , the exponential distribution has mean $1/ λ = 1/2$ . ✓ Matches.

Cell J — exam twist: infinite end AND a spike together

Worked example Example 10 —

$\int_{0}^{\infty} \frac{d x}{x ( 1 + x )}$ Forecast: TWO bad things at once — a spike at $x = 0$ (from $1/ x$ ) AND an infinite upper limit. Split into the two dangers and check each.

Step 1. Split at $x = 1$ (any positive finite point works): the piece $\int_{0}^{1}$ isolates the spike at $0$ , the piece $\int_{1}^{\infty}$ isolates infinity. $\int_{0}^{\infty} = \int_{0}^{1} + \int_{1}^{\infty} .$ Why this step? Each improper feature must be isolated so that every piece carries exactly one danger — this is a combined Type I + Type II integral and the two failures must be tested separately.

Step 2. Find the antiderivative with the substitution $u = x$ , so $x = u^{2}$ and $d x = 2 u d u$ : $\int \frac{d x}{x ( 1 + x )} = \int \frac{2 u d u}{u ( 1 + u ^{2} )} = \int \frac{2 d u}{1 + u ^{2}} = 2 arctan u = 2 arctan x .$ Why this step? The $x$ makes direct integration hard; substituting $u = x$ cancels the $x$ against the $2 u$ from $d x$ , turning the integral into the standard $\frac{1}{1 + u ^{2}}$ pattern whose antiderivative is $arctan$ (the same tool we used in Cell D).

Step 3. Evaluate both pieces with this single antiderivative, taking a limit at each danger: $\int_{0}^{1} = lim_{s \to 0^{+}} [2 arctan x]_{s}^{1} = 2 arctan 1 - lim_{s \to 0^{+}} 2 arctan s = 2 \cdot \frac{π}{4} - 0 = \frac{π}{2},$ $\int_{1}^{\infty} = lim_{t \to \infty} [2 arctan x]_{1}^{t} = lim_{t \to \infty} 2 arctan t - 2 arctan 1 = 2 \cdot \frac{π}{2} - \frac{π}{2} = \frac{π}{2} .$ Why this step? Each piece gets its own movable wall ( $s \to 0^{+}$ for the spike, $t \to \infty$ for infinity); near $0$ the term $2 arctan s \to 0$ (mild spike, finite) and at infinity $2 arctan t \to 2 \cdot \frac{π}{2} = π$ (the angle saturates at $9 0^{\circ}$ ).

Both pieces converge, so total $= \frac{π}{2} + \frac{π}{2} = π$ . Converges to $π$ . Cell J confirmed.

Verify: Overall $2 arctan x$ evaluated from $0$ to $\infty$ is $2 \cdot \frac{π}{2} - 0 = π \approx 3.14159$ , and the two split pieces $\frac{π}{2} + \frac{π}{2}$ agree. ✓ Converges to $π$ .

Recall Quick self-check on the matrix

Which cell is $\int_{- 1}^{1} \frac{d x}{x ^{2}}$ ? ::: Cell G — interior blow-up at $x = 0$ , diverges. Which cell is $\int_{0}^{\infty} e^{- x} d x$ ? ::: Cell C — exponential decay at infinity, converges. Which cell is $\int_{- \infty}^{\infty} x d x$ ? ::: Cell E — signed, principal-value trap, diverges. Which cell is $\int_{0}^{1} x^{- 1/2} d x$ ? ::: Cell F — mild spike at $0$ ( $p < 1$ ), converges.