Improper Integrals. Type II (Vertical Asymptotes)

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Definite Integrals and Improper Integration

A definite integral is a fundamental concept in calculus used to calculate the exact area under a curve between two fixed limits on the x-axis. It is written or represented by the notation: $\int_{a}^{b} f(x)dx$ where:

• $a$ and $b$ are the lower and upper limits of integration.
• $f(x)$ is the integrand (the function being integrated).
• $dx$ indicates the variable of integration.

Conceptually, the definite integral of a real-valued function $f(x)$ on an interval $[a, b]$ is defined as the net signed area of the region bounded by the function's graph, the x-axis, and the vertical lines $x = a$ and $x = b$ and it is illustrated in Figure 1.a.

The Fundamental Theorem of Calculus

The connection between derivatives and integrals is established by the Fundamental Theorem of Calculus. It states that if you have a continuous function $f(x)$ on an closed interval $[a,b]$, the integral of that function is equal to the change in its antiderivative $F(x)$ (where $F'(x) = f(x)$) between the ends of the interval. Mathematically, this is expressed as $\boxed{\int_{a}^{b} f(x)dx = F(b) - F(a) = F(x) \bigg\vert{}_{a}^{b}}$

Improper integration

Standard definite integrals require specific, finite limits of integration and a function that is continuous (and thus bounded) across that entire interval.

Improper integrals break these rules. They arise in two main scenarios:

1. Infinite Intervals: The limits of integration extend to infinity, e.g., $[a, \infty)$, $(-\infty, b]$, or $(-\infty, \infty)$.
2. Infinite Discontinuities (Singularities): The integrand becomes undefined (approaches infinity or negative infinity) at one or more points within the interval of integration.

Because we cannot plug “infinity” directly into an antiderivative, evaluating improper integrals requires rewriting them using limits (as the integration bounds approach infinity or as singularities are approached).

$\boxed{\int_{a}^{\infty} f(x)dx = \lim_{b \to \infty} \int_{a}^{b} f(x)dx}$ (Figure 1.a.).

Improper Integrals: Type II (Vertical Asymptotes)

It is very tempting to calculate integrals blindly using the Fundamental Theorem of Calculus. For example, you might try to evaluate this: $\int_{-1}^{1} \frac{1}{x^2} dx = \left[ -x^{-1} \right]_{-1}^{1} = -1 - (1) = -2 \quad \text{😞😞😞 (WRONG!)}$

Why is this incorrect? Because our area must be positive, yet we got a negative answer. The mistake is that the function $f(x) = \frac{1}{x^2}$ has an infinite discontinuity (a vertical asymptote) at $x = 0$, which lies directly inside our interval $[-1, 1]$.

An improper integral of the second kind arises when the integrand becomes unbounded (approaches $\pm\infty$) either at the endpoints or somewhere within the interval of integration.

Formal Definition

To evaluate these integrals, we replace the troublesome boundary with a variable and take a limit. The integral converges if the limit exists as a finite number, and diverges if it does not.

1. Discontinuity at the lower bound ($a$): $\int_{a}^{b} f(x) dx = \lim_{t \to a^+} \int_{t}^{b} f(x) dx$
2. Discontinuity at the upper bound ($b$): $\int_{a}^{b} f(x) dx = \lim_{t \to b^-} \int_{a}^{t} f(x) dx$
3. Discontinuity in the middle ($c$, where $a < c < b$): $\int_{a}^{b} f(x) dx = \lim_{t \to c^-} \int_{a}^{t} f(x) dx + \lim_{k \to c^+} \int_{k}^{b} f(x) dx$.
   Both limits must converge independently for the entire integral to converge. If either diverges, the whole integral diverges.

The Hierarchy of Asymptotes ($x \to 0^+$)

Before diving into complex examples, let’s look at the baseline $p$-series behavior near zero. As $x$ approaches $0^+$, notice how the functions stack up: $\frac{1}{\sqrt{x}} \ll \frac{1}{x} \ll \frac{1}{x^2}$

• Converges: $\int_{0}^{1} \frac{1}{\sqrt{x}} dx$. The asymptote is “narrow” enough that the area remains finite.
• Diverges: $\int_{0}^{1} \frac{1}{x} dx$ and $\int_{0}^{1} \frac{1}{x^2} dx$. The asymptote is too “wide,” pushing infinite area against the y-axis.

Solved Examples

Endpoint Asymptote (Convergent), $\int_{0}^{1} \frac{1}{\sqrt{x}} dx$

There is a vertical asymptote at $x = 0$.

$\int_{0}^{1} x^{-1/2} dx = \lim_{a \to 0^+} \int_{a}^{1} x^{-1/2} dx = \lim_{a \to 0^+} \left[ 2x^{1/2} \right]_{a}^{1}$

$= \lim_{a \to 0^+} (2 - 2\sqrt{a}) = 2 - 0 = \mathbf{2}$

The improper integral converges to 2 (Figure 1.b.).

Endpoint Asymptote (Divergent), $\int_{0}^{1} \frac{1}{x} dx$

There is an infinite discontinuity at $x = 0$.

$\int_{0}^{1} \frac{1}{x} dx = \lim_{a \to 0^+} \int_{a}^{1} \frac{1}{x} dx = \lim_{a \to 0^+} \left[ \ln\vert{}x\vert{} \right]_{a}^{1}$

$= \lim_{a \to 0^+} (\ln(1) - \ln(a)) = 0 - (-\infty) = \mathbf{\infty}$

This improper integral diverges (Figure 1.c.).

The idea is quite simple, $\frac{1}{x^{1/2}} << \frac{1}{x}<< \frac{1}{x^2}$ as $x \to 0^+$, the first integral is convergent (small enough), but the other two improper integral diverge. Similarly, $\frac{1}{x^{1/2}} >> \frac{1}{x} >> \frac{1}{x^2}$ as $x \to \infty$, the only one that is convergent is the last one.

Partial Fractions with an Asymptote, $\int_{1}^{2} \frac{1}{x^2+x-2} dx$

The denominator factors to $(x-1)(x+2)$, meaning there is an infinite discontinuity at the lower bound $x = 1$.

$\int_{1}^{2} \frac{1}{x^2+x-2} dx = \lim_{a \to 1^+} \int_{a}^{2} \frac{1}{(x-1)(x+2)} dx$

Using partial fractions, we break this into: $\lim_{a \to 1^+} \int_{a}^{2} \left( \frac{1/3}{x-1} - \frac{1/3}{x+2} \right) dx = \lim_{a \to 1^+} \left[ \frac{1}{3}\ln(x-1) - \frac{1}{3}\ln(x+2) \right]_{a}^{2} = \lim_{a \to 1^+} \frac{1}{3} \left[ \ln\left(\frac{x-1}{x+2}\right) \right]_{a}^{2}$

$= \frac{1}{3} \lim_{a \to 1^+} \left( \ln\left(\frac{1}{4}\right) - \ln\left(\frac{a-1}{a+2}\right) \right) = -(-\infty) = \infty$, the integral diverges.

Using U-Substitution

A. Divergent U-Substitution: $\int_{0}^{1} \frac{e^{1/x}}{x^2} dx$

Infinite discontinuity at $x = 0$.

Let $u = \frac{1}{x}$, so $du = -\frac{1}{x^2}dx$. As $x \to a$, $u \to \frac{1}{a}$. As $x \to 1$, $u \to 1$.

$\int_{0}^{1} \frac{e^{1/x}}{x^2} dx = \lim_{a \to 0^+} \int_{a}^{1} \frac{e^{1/x}}{x^2} dx = \lim_{a \to 0^+} \int_{1/a}^{1} -e^u \, du = \lim_{a \to 0^+} \int_{1}^{1/a} e^u \, du = \lim_{a \to 0^+} \left[ e^u \right]_{1}^{1/a}$

$= \lim_{a \to 0^+} \left( e^{1/a} - e \right) = \infty$. The integral diverges.

B. Convergent U-Substitution: $\int_{1}^{e} \frac{1}{x\sqrt{\ln(x)}} dx$

Infinite discontinuity at $x = 1$ because $\ln(1) = 0$.

Let $u = \ln(x)$, $du = \frac{1}{x}dx$. As $x \to a$, $u \to \ln(a)$. As $x \to e$, $u \to 1$.

$\int_{1}^{e} \frac{1}{x\sqrt{\ln(x)}} dx = \lim_{a \to 1^+} \int_{a}^{e} \frac{1}{x\sqrt{\ln(x)}} dx = \lim_{a \to 1^+} \int_{\ln(a)}^{1} \frac{1}{\sqrt{u}} \, du = \lim_{a \to 1^+} \left[ 2\sqrt{u} \right]_{\ln(a)}^{1}$

$= \lim_{a \to 1^+} (2\sqrt{1} - 2\sqrt{\ln(a)}) = 2 - 0 = 2$. The integral converges to $2$.

Discontinuity Inside the Interval, $\int_{0}^{6} \frac{x}{x-2} dx$

Asymptotic behavior occurs at $x = 2$, which is inside the bounds. We must split the integral: $\int_{0}^{6} \frac{x}{x-2} dx =\int_{0}^{2} \frac{x}{x-2} dx + \int_{2}^{6} \frac{x}{x-2} dx$

Testing the first half: $\int_{0}^{2} \frac{x}{x-2} dx = \lim_{b \to 2^-} \int_{0}^{b} \frac{x-2+2}{x-2} dx = \lim_{b \to 2^-} \int_{0}^{b} \left( 1 + \frac{2}{x-2} \right) dx$

$= \lim_{b \to 2^-} \left[ x + 2\ln\vert{}x-2\vert{} \right]_{0}^{b} = \lim_{b \to 2^-} (b + 2\ln\vert{}b-2\vert{}) - (0 + 2\ln\vert{}-2\vert{}) = -\infty$

Because the first half diverges, the entire integral diverges. There is no need to test the second half.

Mixed Improper Integrals (Type 1 & Type 2), $\int_{0}^{\infty} \frac{dx}{(x-3)^2}$

This integral has an infinite bound (Type 1) and an infinite discontinuity at $x=3$ (Type 2). It must be split at the asymptote and at another arbitrary point (e.g., $x=5$): $\int_{0}^{\infty} \frac{dx}{(x-3)^2} = \int_{0}^{3}\frac{dx}{(x-3)^2} + \int_{3}^{5}\frac{dx}{(x-3)^2} + \int_{5}^{\infty}\frac{dx}{(x-3)^2}$

$\int_{5}^{\infty} \frac{dx}{(x-3)^2} = \lim_{b \to \infty} \int_{5}^{b} \frac{dx}{(x-3)^2} = \lim_{b \to \infty} \frac{-1}{x-3}\bigg|_{5}^{b} = $

$\lim_{b \to \infty} (\frac{-1}{b-3}+\frac{1}{5-3}) = \frac{1}{2}$. Hence, $\int_{5}^{\infty} \frac{dx}{(x-3)^2}$ converges, and its value is 1/2.

If we test the segment from $0$ to $3$: $\int_{0}^{3}\frac{dx}{(x-3)^2} = \lim_{c \to 3^-} \int_{0}^{c} \frac{dx}{(x-3)^2} = \lim_{c \to 3^-} \left[ \frac{-1}{x-3} \right]_{0}^{c}$

$= \lim_{c \to 3^-} \left( \frac{-1}{c-3} - \frac{-1}{0-3} \right) = \infty$. Because this single piece diverges, the entire integral diverges (Figure 1.d).

Integration by Parts & L’Hôpital’s Rule, $\int_{0}^{1} x \ln(x) dx$

Vertical asymptote at $x = 0$.

Use integration by parts: $u = \ln(x)$, $dv = x dx \implies du = \frac{1}{x}dx$, $v = \frac{x^2}{2}$.

$\int_{0}^{1} x \ln(x) dx = \lim_{a \to 0^+} \int_{a}^{1} x \ln(x) dx = \lim_{a \to 0^+} \left( \left[ \frac{x^2\ln(x)}{2} \right]_{a}^{1} - \int_{a}^{1} \frac{x}{2} dx \right)$

$= \lim_{a \to 0^+} \left( \left[ \frac{x^2\ln(x)}{2} \right]_{a}^{1} - \left[ \frac{x^2}{4} \right]_{a}^{1} \right)$

Evaluating at the bounds: $= \lim_{a \to 0^+} \left( \frac{1^2\ln(1)}{2} - \frac{1}{4} - \frac{a^2\ln(a)}{2} + \frac{a^2}{4} \right)= \lim_{a \to 0^+} \left( -\frac{1}{4} - \frac{a^2\ln(a)}{2} + \frac{a^2}{4} \right)$

To evaluate the limit of $a^2 \ln(a)$, rewrite it to use L’Hôpital’s Rule: $\lim_{a \to 0^+} \frac{\ln(a)}{2/a^2} \xrightarrow{L'H} \lim_{a \to 0^+} \frac{1/a}{-4/a^3} = \lim_{a \to 0^+} \frac{-a^2}{4} = 0$

Returning to our limit: $\lim_{a \to 0^+} \left( -\frac{1}{4} - \frac{a^2\ln(a)}{2} + \frac{a^2}{4} \right)= -\frac{1}{4} - 0 + 0 = -\frac{1}{4}$. The integral converges to $-1/4$.

Rational Function using Partial Fractions, $\int_{1}^{\infty} \frac{1}{x^2+x} dx$

1. This is a Type I improper integral (it has an infinite limit of integration). To solve it, we first need to break down the rational integrand using partial fraction decomposition.
2. Partial Fraction Decomposition.
   Factor the denominator and set up the decomposition: $\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}$
   Multiply both sides by the common denominator $x(x+1)$: $1 = A(x+1) + Bx$
   To find the coefficients:
   Let $x = 0 \implies 1 = A(1) + 0 \implies \mathbf{A = 1}$
   Let $x = -1 \implies 1 = 0 + B(-1) \implies \mathbf{B = -1}$
3. Set up the Limit Integral.
   Rewrite the integral using our partial fractions and replace the infinite upper bound with a variable $b$: $\int_{1}^{\infty} \frac{1}{x^2+x} dx = \int_{1}^{\infty} \frac{1}{x} dx - \int_{1}^{\infty} \frac{1}{x+1} dx = \lim_{b \to \infty} \int_{1}^{b} \left( \frac{1}{x} - \frac{1}{x+1} \right) dx$
4. Integrate and Evaluate.
   Integrate both terms yielding natural logarithms: $= \lim_{b \to \infty} \left[ \ln\vert{}x\vert{} - \ln\vert{}x+1\vert{} \right]_{1}^{b}$
   Apply the limits of integration (Fundamental Theorem of Calculus): $= \lim_{b \to \infty} \left( (\ln\vert{}b\vert{} - \ln\vert{}b+1\vert{}) - (\ln\vert{}1\vert{} - \ln\vert{}2\vert{}) \right)$
   Use the logarithm quotient property ($\ln(u) - \ln(v) = \ln(u/v)$), $b \gt 0, b + 1 \gt 0$, and the fact that $\ln(1) = 0$: $= \lim_{b \to \infty} \ln\left(\frac{b}{b+1}\right) + \ln(2)$
   As $b \to \infty$, the fraction $\frac{b}{b+1}$ approaches $1$. Since $\ln(1) = 0$: $= 0 + \ln(2) = \mathbf{\ln(2)}$
5. Conclusion: The integral converges to $\ln(2)$, $\int_{1}^{\infty} \frac{1}{x^2+x} dx = \ln(2)$.

Mixed Improper Integral (Type I and Type II), $\int_{2}^{\infty} \frac{1}{x\sqrt{x^2-4}} dx$

1. This integral is “doubly improper”. It has an infinite upper bound (Type I), AND the integrand is undefined at the lower bound $x = 2$ because it causes a zero in the denominator (Type II).
2. Find the Antiderivative
   First, let’s find the indefinite integral. We can manipulate the expression inside the square root to match the standard form for the inverse secant derivative, which is $\int \frac{1}{x\sqrt{x^2-1}}dx = \sec^{-1}(x)$.
   Factor out a $4$ from inside the square root: $\int \frac{1}{x\sqrt{4(\frac{x^2}{4}-1)}} dx = \int \frac{1}{2x\sqrt{(\frac{x}{2})^2-1}} dx$
   Use U-substitution. Let $u = \frac{x}{2}$, so $du = \frac{1}{2} dx$ (which means $dx = 2 du$): $\int \frac{1}{2(2u)\sqrt{u^2-1}} (2 du) = \frac{1}{2} \int \frac{1}{u\sqrt{u^2-1}} du$
   Using the standard inverse trigonometric formula: $\int \frac{1}{u\sqrt{u^2-1}} du = \frac{1}{2} \sec^{-1}(u) = \mathbf{\frac{1}{2} \sec^{-1}\left(\frac{x}{2}\right)}$
3. Split the Improper Integral
   Because the discontinuity is at a boundary, we must split the integral at an arbitrary point $c > 2$ (let’s use $x = 3$) to separate the two types of improper limits: $\int_{2}^{\infty} \frac{1}{x\sqrt{x^2-4}} dx = \int_{2}^{3} \frac{1}{x\sqrt{x^2-4}} dx + \int_{3}^{\infty} \frac{1}{x\sqrt{x^2-4}} dx$
4. Evaluate the First Part (Type II)
   Let a variable approach the discontinuity at $x=2$ from the right: $\lim_{a \to 2^+} \int_{a}^{3} \frac{1}{x\sqrt{x^2-4}} dx = \lim_{a \to 2^+} \left[ \frac{1}{2} \sec^{-1}\left(\frac{x}{2}\right) \right]_{a}^{3}$
   $= \lim_{a \to 2^+} \left( \frac{1}{2} \sec^{-1}\left(\frac{3}{2}\right) - \frac{1}{2} \sec^{-1}\left(\frac{a}{2}\right) \right)$
   As $a \to 2^+$, $\frac{a}{2} \to 1$. We know that $\sec^{-1}(1) = 0$: $= \frac{1}{2} \sec^{-1}\left(\frac{3}{2}\right) - \frac{1}{2}(0) = \mathbf{\frac{1}{2} \sec^{-1}\left(\frac{3}{2}\right)}$
5. Evaluate the Second Part (Type I)
   Let a variable approach infinity: $\lim_{b \to \infty} \int_{3}^{b} \frac{1}{x\sqrt{x^2-4}} dx = \lim_{b \to \infty} \left( \frac{1}{2}\sec^{-1}\left(\frac{b}{2}\right) - \frac{1}{2}\sec^{-1}\left(\frac{3}{2}\right) \right)$
   Visualizing the Limit: What happens to $\sec^{-1}(\frac{b}{2})$ as $b$ grows infinitely large? The function $\sec(x)$ goes to infinity as $x$ approaches $\frac{\pi}{2}$ from the left.
   Therefore, its inverse function, $\sec^{-1}(x)$, approaches a horizontal asymptote of $\frac{\pi}{2}$ as $x \to \infty$.
   Knowing this limit, we can finish evaluating the second part: $= \frac{1}{2} \left( \frac{\pi}{2} \right) - \frac{1}{2}\sec^{-1}\left(\frac{3}{2}\right) = \mathbf{\frac{\pi}{4} - \frac{1}{2}\sec^{-1}\left(\frac{3}{2}\right)}$
6. Combine the Results
   Add the two evaluated limits together to find the final area: $\int_{2}^{\infty} \frac{1}{x\sqrt{x^2-4}} dx = \left[ \frac{1}{2} \sec^{-1}\left(\frac{3}{2}\right) \right] + \left[ \frac{\pi}{4} - \frac{1}{2}\sec^{-1}\left(\frac{3}{2}\right) \right]$
   The $\sec^{-1}(\frac{3}{2})$ terms perfectly cancel out, leaving us with: $= \mathbf{\frac{\pi}{4}}$
7. Conclusion: The integral converges to $\frac{\pi}{4}$.

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