Helpful context:

Here is the normal distribution’s density function:

$$f(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}.$$

For $f$ to be a valid probability density, the total area under the curve must equal $1$:

$$\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-x^2/2} dx = 1.$$

But wait. The Riemann integral we built in the previous post was defined on a closed, bounded interval $[a, b]$. This integral has limits $-\infty$ and $\infty$. The interval is all of $\mathbb{R}$. It is not bounded. The Riemann definition, as we stated it, does not apply.

And yet - the question is real. For the normal distribution to make sense as a probability distribution, this infinite integral must equal $1$. Whether it does is not a philosophical question. It is a mathematical question that requires an answer.

This is the problem of improper integrals: extending the theory of integration to cover infinite domains and functions that blow up somewhere in the domain. The extension is elegant, surprisingly tractable, and absolutely necessary for probability theory.

Section 1: What “Improper” Means

The Riemann integral $\int_a^b f(x) dx$ requires:

  1. A bounded interval $[a, b]$.
  2. A bounded function $f$ on that interval.

An integral is improper if either condition fails:

  • Type 1: one or both limits are infinite ($a = -\infty$ or $b = \infty$ or both).
  • Type 2: $f$ has a vertical asymptote somewhere in $[a, b]$ (the function blows up).

Discomfort check. Why is a new definition necessary? Why not just say the integral is a very large number and move on? Because the Riemann integral is built on the idea of partitioning a finite interval into finitely many subintervals. There is no way to partition $[1, \infty)$ into finitely many subintervals and have them cover the whole domain. The mathematical object $\int_1^\infty f(x) dx$ does not yet have meaning - we need to define what we mean by it before we can ask whether it equals anything. The definition as a limit is not a workaround. It is the correct thing.

The fix in both cases is the same: replace the problematic limit with a variable, compute the ordinary Riemann integral, then take a limit.

What these integrals intuitively mean. The definition-as-limit is not just a technical patch - it captures a genuine question about accumulation.

Type 1 (infinite limits). The integral $\int_a^\infty f(x) dx$ is asking: if you accumulate area starting at $a$ and keep pushing the right boundary further and further out - to $10$, to $10^6$, to $10^{12}$ - does the running total eventually settle at a fixed number, or does it keep climbing without bound? This is exactly the question “does this infinite series converge?” wearing different clothes. If $f(x)$ shrinks fast enough, each new strip you add contributes so little that the total barely moves, and the running total converges. If $f(x)$ shrinks too slowly, the strips keep contributing meaningfully and the total grows forever. The limit $\lim_{b \to \infty} \int_a^b f(x) dx$ is just the mathematical way of asking whether that running total stabilizes.

Type 2 (vertical asymptotes). Near a blow-up point $x = a$, the strips are extremely tall - the function is huge - but they are also extremely narrow, since $dx \to 0$ as $x \to a^+$. This is a race: height grows, width shrinks. The area of each strip is approximately $f(x) \cdot dx$. Whether the total area is finite depends on which effect wins. If width shrinks faster than height grows, the spike contributes a finite total area - the blow-up is integrable. If height grows faster than width shrinks, the spike contributes infinite area - the function is too singular. The $p$-integral (Section 4) makes this race precise: $1/x^p$ near $0$ converges for $p < 1$ (width wins) and diverges for $p \geq 1$ (height wins).

Section 2: Type 1 - Infinite Limits

Definition. If $f$ is integrable on $[a, b]$ for every $b > a$, we define:

$$\int_a^\infty f(x) dx = \lim_{b \to \infty} \int_a^b f(x) dx,$$

provided this limit exists and is finite. If it does, we say the integral converges. If not, it diverges.

Similarly: $\int_{-\infty}^b f(x) dx = \lim_{a \to -\infty} \int_a^b f(x) dx$.

For a doubly-infinite integral, split at any convenient point $c$:

$$\int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^c f(x) dx + \int_c^\infty f(x) dx,$$

and both pieces must converge separately.

The canonical example. $\int_1^\infty \frac{1}{x^2} dx$.

$$\int_1^b \frac{1}{x^2} dx = \left[-\frac{1}{x}\right]_1^b = -\frac{1}{b} + 1 = 1 - \frac{1}{b}.$$

As $b \to \infty$: $1 - \frac{1}{b} \to 1$.

$$\int_1^\infty \frac{1}{x^2} dx = 1.$$

The area under $1/x^2$ from $1$ to $\infty$ is exactly $1$. This is a finite region - the curve falls fast enough that infinitely many thin strips add up to a finite total.

The contrasting example. $\int_1^\infty \frac{1}{x} dx$.

$$\int_1^b \frac{1}{x} dx = [\ln x]_1^b = \ln b - \ln 1 = \ln b.$$

As $b \to \infty$: $\ln b \to \infty$.

$$\int_1^\infty \frac{1}{x} dx = \infty. \quad \text{(diverges)}$$

The area under $1/x$ from $1$ to $\infty$ is infinite.

Both $1/x^2$ and $1/x$ approach $0$ as $x \to \infty$. Both functions become arbitrarily small. And yet one gives finite area and the other infinite area. This is the key tension of improper integrals, and it deserves to be felt as genuinely surprising.

Discomfort check. How can two functions that both approach $0$ give such different results? The answer is rate. The function $1/x^2$ falls much faster than $1/x$. At $x = 100$: $1/x^2 = 0.0001$ while $1/x = 0.01$, already a factor of $100$ different. The strips under $1/x^2$ shrink fast enough that their sum converges. The strips under $1/x$ shrink, but not fast enough - they contribute roughly $\ln b$ total area out to $b$, and $\ln b$ grows without bound. The intuition is: going to zero is necessary for convergence, but not sufficient. Going to zero fast enough is what matters.

To make “fast enough” precise, we need the $p$-integral.

Section 3: The $p$-Integral

The family $\int_1^\infty \frac{1}{x^p} dx$ for various $p > 0$ is the central test case for improper integrals. Understanding it gives you a complete picture of the rate-versus-convergence trade-off.

Case 1: $p \neq 1$.

$$\int_1^b \frac{1}{x^p} dx = \left[\frac{x^{1-p}}{1-p}\right]_1^b = \frac{b^{1-p}}{1-p} - \frac{1}{1-p} = \frac{b^{1-p} - 1}{1-p}.$$

Now take $b \to \infty$:

  • If $p > 1$: the exponent $1 - p < 0$, so $b^{1-p} = b^{-(p-1)} \to 0$. The limit is $\frac{0 - 1}{1-p} = \frac{1}{p-1}$.
  • If $p < 1$: the exponent $1 - p > 0$, so $b^{1-p} \to \infty$. The integral diverges.

Case 2: $p = 1$. We already computed: $\int_1^b \frac{1}{x} dx = \ln b \to \infty$. Diverges.

Theorem ($p$-integral).

$$\int_1^\infty \frac{1}{x^p} dx = \begin{cases} \dfrac{1}{p-1} & \text{if } p > 1 \\ \infty & \text{if } p \leq 1. \end{cases}$$

The cutoff is at $p = 1$ exactly. Powers greater than $1$ give convergence; powers $1$ or less give divergence.

Discomfort check. Why is the cutoff at exactly $p = 1$? The short answer is the algebra above: for $p > 1$, the antiderivative $\frac{x^{1-p}}{1-p}$ has a negative exponent, making $b^{1-p} \to 0$ as $b \to \infty$. For $p < 1$, the antiderivative has a positive exponent, making $b^{1-p} \to \infty$. At $p = 1$ exactly, the antiderivative is $\ln x$, which grows without bound but more slowly than any positive power. The deeper reason: power functions $x^p$ for $p > 1$ decrease fast enough that the “tail” from $1$ to $\infty$ is summable; for $p \leq 1$, the tail is too heavy. The value $p = 1$ is the phase boundary - the exact threshold where the behavior changes. This kind of sharp threshold appears throughout analysis and is worth remembering as a structural fact: power laws near $1$ are the critical case.

Section 4: Type 2 - Vertical Asymptotes

The second type of improper integral arises when the integrand has a vertical asymptote somewhere in the domain.

Definition. If $f$ has a vertical asymptote at $x = a$ but is integrable on $[a + \varepsilon, b]$ for every $\varepsilon > 0$, define:

$$\int_a^b f(x) dx = \lim_{\varepsilon \to 0^+} \int_{a+\varepsilon}^b f(x) dx,$$

if this limit exists and is finite.

Similarly, for an asymptote at $x = b$:

$$\int_a^b f(x) dx = \lim_{\varepsilon \to 0^+} \int_a^{b-\varepsilon} f(x) dx.$$

If the asymptote is at an interior point $c \in (a, b)$, split the integral at $c$:

$$\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx,$$

and both pieces must converge.

Example. $\int_0^1 \frac{1}{\sqrt{x}} dx$.

The function $\frac{1}{\sqrt{x}} \to \infty$ as $x \to 0^+$. Using the definition:

$$\int_\varepsilon^1 \frac{1}{\sqrt{x}} dx = \left[2\sqrt{x}\right]_\varepsilon^1 = 2 - 2\sqrt{\varepsilon}.$$

As $\varepsilon \to 0^+$: $2 - 2\sqrt{\varepsilon} \to 2$.

$$\int_0^1 \frac{1}{\sqrt{x}} dx = 2.$$

The function blows up at $0$, but gently enough that the area under the spike is finite.

Contrasting example. $\int_0^1 \frac{1}{x} dx$.

$$\int_\varepsilon^1 \frac{1}{x} dx = \ln 1 - \ln \varepsilon = -\ln \varepsilon.$$

As $\varepsilon \to 0^+$: $-\ln \varepsilon \to \infty$. Diverges.

The $p$-integral near a singularity. The analog of the $p$-test applies to $\int_0^1 \frac{1}{x^p} dx$:

$$\int_0^1 \frac{1}{x^p} dx = \begin{cases} \dfrac{1}{1-p} & \text{if } p < 1 \\ \infty & \text{if } p \geq 1. \end{cases}$$

Note the reversal: for singularities at the left endpoint, convergence requires $p < 1$, not $p > 1$. The physics is the same - the singularity must not be too severe - but the direction of inequality flips because we are looking at $x \to 0$ instead of $x \to \infty$.

Section 5: The Comparison Test

Most improper integrals we encounter in practice cannot be computed exactly. Instead of computing, we compare.

Theorem (Direct Comparison Test). Let $0 \leq f(x) \leq g(x)$ for all $x \geq a$.

  • If $\int_a^\infty g(x) dx$ converges, then $\int_a^\infty f(x) dx$ converges (and is $\leq \int_a^\infty g(x) dx$).
  • If $\int_a^\infty f(x) dx$ diverges, then $\int_a^\infty g(x) dx$ diverges.

Proof. Let $F(b) = \int_a^b f(x) dx$ and $G(b) = \int_a^b g(x) dx$. Since $f, g \geq 0$, both $F$ and $G$ are increasing functions of $b$.

If $\int_a^\infty g dx = L < \infty$: then $F(b) \leq G(b) \leq L$ for all $b$. An increasing function bounded above converges (by the monotone convergence principle, a consequence of the completeness of $\mathbb{R}$). So $\int_a^\infty f dx$ converges.

If $\int_a^\infty f dx = \infty$: then $G(b) \geq F(b) \to \infty$, so $\int_a^\infty g dx = \infty$. $\blacksquare$

The intuition: if the bigger function has finite area, the smaller one certainly does. If the smaller one has infinite area, the bigger one has even more.

Example. Does $\int_1^\infty \frac{1}{x^2 + 1} dx$ converge?

For $x \geq 1$: $x^2 + 1 \geq x^2$, so $\frac{1}{x^2 + 1} \leq \frac{1}{x^2}$.

Since $\int_1^\infty \frac{1}{x^2} dx = 1$ converges, the comparison test says $\int_1^\infty \frac{1}{x^2 + 1} dx$ also converges.

(In this case we could compute exactly: $\int_1^\infty \frac{1}{x^2+1} dx = [\arctan x]_1^\infty = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$. But the point is that comparison tells us it converges even before we compute.)

Example. Does $\int_1^\infty \frac{x + 1}{x^3 - x + 1} dx$ converge?

For large $x$: $x^3 - x + 1 \approx x^3$, so $\frac{x+1}{x^3 - x + 1} \approx \frac{x}{x^3} = \frac{1}{x^2}$.

More precisely, for $x \geq 2$: $x^3 - x + 1 \geq \frac{x^3}{2}$ (since $x \leq x^3/4$ for large $x$, and $-1 \leq 0$). And $x + 1 \leq 2x$. So:

$$\frac{x+1}{x^3 - x + 1} \leq \frac{2x}{x^3/2} = \frac{4}{x^2}.$$

Since $\int_2^\infty \frac{4}{x^2} dx$ converges (it is $4 \cdot \frac{1}{1} = 4$), the original integral converges.

Section 6: The Limit Comparison Test

The direct comparison test requires finding an explicit bound. A more flexible tool is the limit comparison test.

Theorem (Limit Comparison Test). Let $f, g \geq 0$ on $[a, \infty)$. If:

$$\lim_{x \to \infty} \frac{f(x)}{g(x)} = L, \quad 0 < L < \infty,$$

then $\int_a^\infty f(x) dx$ and $\int_a^\infty g(x) dx$ either both converge or both diverge.

Proof sketch. Since $f/g \to L$, for large $x$ (say $x > M$) the ratio is between $L/2$ and $2L$:

$$\frac{L}{2} g(x) \leq f(x) \leq 2L g(x) \quad \text{for } x > M.$$

If $\int_M^\infty g dx$ converges, then $\int_M^\infty f dx \leq 2L \int_M^\infty g dx < \infty$ (converges). If $\int_M^\infty g dx$ diverges, then $\int_M^\infty f dx \geq \frac{L}{2} \int_M^\infty g dx = \infty$ (diverges). The finite portion $\int_a^M$ is a proper integral and does not affect convergence. $\blacksquare$

Example. Does $\int_1^\infty \frac{x^2 + 3}{x^4 + 2x + 1} dx$ converge?

Compare with $g(x) = \frac{1}{x^2}$:

$$\frac{f(x)}{g(x)} = \frac{x^2(x^2 + 3)}{x^4 + 2x + 1} = \frac{x^4 + 3x^2}{x^4 + 2x + 1} \to 1 \quad \text{as } x \to \infty.$$

Since the limit is $1$ (finite and nonzero) and $\int_1^\infty \frac{1}{x^2} dx$ converges, our integral converges.

Example. Does $\int_2^\infty \frac{1}{\sqrt{x^2 - 1}} dx$ converge?

For large $x$: $\sqrt{x^2 - 1} \approx \sqrt{x^2} = x$. Compare with $g(x) = 1/x$:

$$\frac{f(x)}{g(x)} = \frac{x}{\sqrt{x^2-1}} \to 1.$$

Since $\int_2^\infty \frac{1}{x} dx$ diverges, our integral also diverges.

Section 7: The Gaussian Integral

Here is the most important improper integral in mathematics.

$$\int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi}.$$

This integral cannot be computed by finding an elementary antiderivative - $e^{-x^2}$ has none. Instead, it requires a trick: convert to a double integral, switch to polar coordinates.

The trick. Let $I = \int_{-\infty}^\infty e^{-x^2} dx$. Then:

$$I^2 = \left(\int_{-\infty}^\infty e^{-x^2} dx\right)\left(\int_{-\infty}^\infty e^{-y^2} dy\right) = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2 + y^2)} dx dy.$$

The integrand $e^{-(x^2+y^2)}$ depends only on the distance from the origin, $r = \sqrt{x^2 + y^2}$. Switch to polar coordinates: $x = r\cos\theta$, $y = r\sin\theta$, and $dx dy = r dr d\theta$:

$$I^2 = \int_0^{2\pi} \int_0^\infty e^{-r^2} r dr d\theta = 2\pi \int_0^\infty e^{-r^2} r dr.$$

Now the inner integral is tractable. Substitute $u = r^2$, $du = 2r dr$:

$$\int_0^\infty e^{-r^2} r dr = \frac{1}{2}\int_0^\infty e^{-u} du = \frac{1}{2}\left[-e^{-u}\right]_0^\infty = \frac{1}{2}(0 - (-1)) = \frac{1}{2}.$$

Therefore $I^2 = 2\pi \cdot \frac{1}{2} = \pi$, so $I = \sqrt{\pi}$.

$$\int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi}.$$

By the substitution $x \to x/\sqrt{2}$ (so $x^2 \to x^2/2$ and $dx \to dx/\sqrt{2}$):

$$\int_{-\infty}^\infty e^{-x^2/2} dx = \sqrt{2\pi}.$$

Dividing both sides by $\sqrt{2\pi}$:

$$\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} e^{-x^2/2} dx = 1.$$

This is why the normal distribution has the normalizing constant $\sqrt{2\pi}$ in its denominator. The constant is not chosen for aesthetic reasons or computational convenience. It is the unique constant that makes the density integrate to $1$. The existence of that constant - and its precise value $\sqrt{2\pi}$ - is a consequence of the Gaussian integral.

Section 8: Probability as Improper Integration

This section makes explicit why improper integrals underlie all of continuous probability.

A probability density function is a function $f: \mathbb{R} \to \mathbb{R}$ with:

  1. $f(x) \geq 0$ for all $x$.
  2. $\int_{-\infty}^\infty f(x) dx = 1$.

Every single continuous probability distribution is defined by specifying such a function. The integral in condition (2) is always an improper integral (limits are $\pm \infty$). The distribution is valid - it represents a genuine probability measure - if and only if this improper integral converges to $1$.

Verifying specific distributions:

The exponential distribution has $f(x) = \lambda e^{-\lambda x}$ for $x \geq 0$ and $f(x) = 0$ for $x < 0$, with $\lambda > 0$.

$$\int_0^\infty \lambda e^{-\lambda x} dx = \lambda \left[-\frac{1}{\lambda} e^{-\lambda x}\right]_0^\infty = [-e^{-\lambda x}]_0^\infty = 0 - (-1) = 1.$$

The Cauchy distribution has density $f(x) = \frac{1}{\pi(1 + x^2)}$.

$$\int_{-\infty}^\infty \frac{1}{\pi(1+x^2)} dx = \frac{1}{\pi}[\arctan x]_{-\infty}^\infty = \frac{1}{\pi}\left(\frac{\pi}{2} - \left(-\frac{\pi}{2}\right)\right) = 1.$$

The Cauchy distribution is a famous example of a distribution with no finite expected value:

$$E[X] = \int_{-\infty}^\infty \frac{x}{\pi(1+x^2)} dx$$

This integral diverges (both the positive and negative parts diverge separately). The Cauchy distribution is well-defined as a probability distribution, but it has no mean. It is heavy-tailed enough that the average of Cauchy samples does not converge as the sample size grows.

Expected values. For a random variable with density $f$:

$$E[X] = \int_{-\infty}^\infty x f(x) dx.$$

This integral may or may not converge. When it does, the expected value exists. When it diverges, the distribution has no finite mean.

Variance. $\text{Var}(X) = \int_{-\infty}^\infty (x - \mu)^2 f(x) dx$. Again, an improper integral that may or may not converge.

The existence of moments (mean, variance, and higher moments) is always an improper integral convergence question. Distributions with heavy tails - Cauchy, Pareto, and certain power-law distributions - can fail to have finite variance or even finite mean, and this failure has real statistical consequences (sample means do not converge, central limit theorem fails to apply in the standard form).

Section 9: The Gamma Function

The $p$-integral at a singularity gives rise to one of the most important functions in mathematics.

Definition. The Gamma function is defined for $s > 0$ by:

$$\Gamma(s) = \int_0^\infty x^{s-1} e^{-x} dx.$$

This integral is improper in two ways: the upper limit is $\infty$, and for $0 < s < 1$ the integrand blows up at $x = 0$ (since $x^{s-1} \to \infty$ as $x \to 0^+$ when $s < 1$). Both issues are resolved by limits.

Convergence. Near $x = 0$: $e^{-x} \approx 1$, so the integrand behaves like $x^{s-1}$. By the $p$-test for singularities (with $p = 1 - s$), the integral near $0$ converges if $1 - s < 1$, i.e., $s > 0$. Near $x = \infty$: the exponential $e^{-x}$ decays faster than any polynomial, so the tail converges easily.

Key property. Integration by parts gives $\Gamma(s+1) = s \cdot \Gamma(s)$.

Proof. Integrate $\int_0^\infty x^s e^{-x} dx$ by parts with $u = x^s$ and $dv = e^{-x} dx$:

$$\Gamma(s+1) = [-x^s e^{-x}]_0^\infty + s \int_0^\infty x^{s-1} e^{-x} dx = 0 + s \cdot \Gamma(s) = s \cdot \Gamma(s). \quad \blacksquare$$

Base case. $\Gamma(1) = \int_0^\infty e^{-x} dx = 1$.

The recurrence $\Gamma(s+1) = s \cdot \Gamma(s)$ and the base case give:

$$\Gamma(n) = (n-1)! \quad \text{for positive integers } n.$$

So the Gamma function is a continuous extension of the factorial to non-integer arguments. For example, $\Gamma(1/2) = \sqrt{\pi}$ (this follows from the Gaussian integral, by a substitution $x = t^2$).

The Gamma function appears in the normalizing constants of the chi-squared distribution, the $t$-distribution, the beta distribution, and many others used in statistics. Every time you see a complicated normalizing constant in a statistical distribution, there is likely a Gamma function behind it.

Section 10: Conditional Convergence

An improper integral of a function that takes both positive and negative values can converge even if the integral of the absolute value diverges. This is called conditional convergence and requires care.

Example. The integral $\int_1^\infty \frac{\sin x}{x} dx$ converges (this is a delicate result, provable by integration by parts). But $\int_1^\infty \frac{|\sin x|}{x} dx$ diverges (comparable to $\int_1^\infty \frac{1}{x} dx$ on intervals where $|\sin x|$ is bounded below).

The comparison and limit comparison tests apply only to nonnegative functions. For functions of variable sign, convergence is more subtle.

For absolutely convergent integrals - those where $\int |f|$ converges - the convergence is robust: you can rearrange the order of summation (in the Riemann sum approximation) without affecting the result. For conditionally convergent integrals, rearrangement can change the value. This is the integral analog of the rearrangement problems for conditionally convergent series.

In probability and statistics, we mostly deal with densities $f \geq 0$, so conditional convergence is not the main concern. But in Fourier analysis, the integral $\int_{-\infty}^\infty f(x) e^{-2\pi i \xi x} dx$ involves the complex exponential, which oscillates and takes both positive and negative real parts. Convergence of the Fourier transform is a conditional convergence question. This is why the Fourier transform requires careful function-space assumptions.

Section 11: Summary of Convergence Tests

Here is a practical guide for determining whether an improper integral converges.

Step 1: Identify the problem. Is the issue an infinite limit (Type 1) or a vertical asymptote (Type 2) - or both?

Step 2: For simple cases, compute directly. If you can find an antiderivative, compute the limit explicitly.

Step 3: For Type 1 ($\int_a^\infty f dx$) with $f \geq 0$: compare with $1/x^p$.

  • If $f(x) \leq C/x^p$ for $p > 1$ and large $x$: converges.
  • If $f(x) \geq C/x^p$ for $p \leq 1$ and large $x$: diverges.
  • For asymptotic comparisons: use limit comparison with $1/x^p$ for the $p$ suggested by the leading behavior.

Step 4: For Type 2 ($\int_a^b f dx$ with asymptote at $a$) with $f \geq 0$: compare with $1/x^p$ near the singularity.

  • If $f(x) \leq C/(x-a)^p$ for $p < 1$: converges.
  • If $f(x) \geq C/(x-a)^p$ for $p \geq 1$: diverges.

Step 5: For functions of variable sign: consider $\int |f|$. If that converges, you have absolute convergence. If not, use more specialized tools.

Example with both types. $\int_0^\infty \frac{1}{\sqrt{x}(1+x)} dx$ has a singularity at $x = 0$ (Type 2) and an infinite upper limit (Type 1). Split at $x = 1$:

  • Near $x = 0$: $\frac{1}{\sqrt{x}(1+x)} \approx \frac{1}{\sqrt{x}}$. Since $p = 1/2 < 1$, $\int_0^1 \frac{1}{\sqrt{x}} dx$ converges. So the left piece converges.
  • Near $x = \infty$: $\frac{1}{\sqrt{x}(1+x)} \approx \frac{1}{x^{3/2}}$. Since $p = 3/2 > 1$, $\int_1^\infty \frac{1}{x^{3/2}} dx$ converges. So the right piece converges.

Both pieces converge, so the full integral converges. (Its exact value is $\pi$, computable via residues or substitution $x = t^2$ followed by partial fractions.)

Summary

Concept Statement
Type 1 improper integral $\int_a^\infty f = \lim_{b \to \infty} \int_a^b f$; converges if limit is finite
Type 2 improper integral $\int_a^b f = \lim_{\varepsilon \to 0^+} \int_{a+\varepsilon}^b f$ when $f$ blows up at $a$
$p$-integral (Type 1) $\int_1^\infty x^{-p} dx$ converges iff $p > 1$; value is $\frac{1}{p-1}$
$p$-integral (Type 2) $\int_0^1 x^{-p} dx$ converges iff $p < 1$; value is $\frac{1}{1-p}$
Direct comparison $0 \leq f \leq g$: if $\int g < \infty$ then $\int f < \infty$; if $\int f = \infty$ then $\int g = \infty$
Limit comparison $f/g \to L \in (0, \infty)$: $\int f$ and $\int g$ converge or diverge together
Gaussian integral $\int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi}$; computed via polar coordinates
Normal distribution Normalizing constant $\sqrt{2\pi}$ comes directly from the Gaussian integral
Probability densities $\int_{-\infty}^\infty f(x) dx = 1$ is always an improper integral convergence statement
Gamma function $\Gamma(s) = \int_0^\infty x^{s-1} e^{-x} dx$; extends factorial to non-integers

The theory of improper integrals is not a technical annex to integration. It is the part of integration that makes probability theory work. Every density function is defined by an improper integral. Every moment (mean, variance) is an improper integral. The question of whether a distribution has a finite mean is a convergence question. Understanding improper integrals is understanding the mathematical foundation on which all of continuous probability rests.

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