Reading guide · Proof index

Improper integrals

Jiří Lebl, Basic Analysis I–II, version 6.3. Free author edition of this section. Selection and attribution · Notation.

L5.5.3: Complete scalar tail identity for an infinite right endpoint.

Proof.

Let c>b.c > b\text{.} Then
∫acf=∫abf+∫bcf.\begin{equation*} \int_a^c f = \int_a^b f + \int_b^c f . \end{equation*}
Taking the limit c→∞c \to \infty finishes the proof.

L5.5.4: Complete nonnegative-supremum and divergent-endpoint subsequence proof.

Proof.

We start with the first item. As ff is nonnegative, ∫axf\int_a^x f is increasing as a function of x.x\text{.} If the supremum is infinite, then for every M∈RM \in \R we find NN such that ∫aNf≥M.\int_a^N f \geq M\text{.} As ∫axf\int_a^x f is increasing, ∫axf≥M\int_a^x f \geq M for all x≥N.x \geq N\text{.} So ∫a∞f\int_a^\infty f diverges to infinity.
Next suppose the supremum is finite, say A≔sup⁡{∫axf:x≥a}.A \coloneqq \sup \left\{ \int_a^x f : x \geq a \right\}\text{.} For every ϵ>0,\epsilon > 0\text{,} we find an NN such that A−∫aNf<ϵ.A - \int_a^N f < \epsilon\text{.} As ∫axf\int_a^x f is increasing, then A−∫axf<ϵA - \int_a^x f < \epsilon for all x≥Nx \geq N and hence ∫a∞f\int_a^\infty f converges to A.A\text{.}
Let us look at the second item. If ∫a∞f\int_a^\infty f converges, then every sequence {xn}n=1∞\{ x_n \}_{n=1}^\infty going to infinity works. The trick is proving the other direction. Suppose {xn}n=1∞\{ x_n \}_{n=1}^\infty is such that lim⁡n→∞xn=∞\lim_{n\to\infty} x_n = \infty and
lim⁡n→∞∫axnf=A\begin{equation*} \lim_{n\to\infty} \int_a^{x_n} f = A \end{equation*}
converges. Given ϵ>0,\epsilon > 0\text{,} pick NN such that for all n≥N,n \geq N\text{,} we have A−ϵ<∫axnf<A+ϵ.A - \epsilon < \int_a^{x_n} f < A + \epsilon\text{.} Because ∫axf\int_a^x f is increasing as a function of x,x\text{,} we have that for all x≥xNx \geq x_N
A−ϵ<∫axNf≤∫axf.\begin{equation*} A - \epsilon < \int_a^{x_N} f \leq \int_a^x f . \end{equation*}
As {xn}n=1∞\{ x_n \}_{n=1}^\infty goes to ∞,\infty\text{,} we have that for any x,x\text{,} there is an xmx_m such that m≥Nm \geq N and x≤xm.x \leq x_m\text{.} Then
∫axf≤∫axmf<A+ϵ.\begin{equation*} \int_a^{x} f \leq \int_a^{x_m} f < A + \epsilon . \end{equation*}
In particular, for all x≥xN,x \geq x_N\text{,} we have ∣∫axf−A∣<ϵ.\abs{\int_a^{x} f - A} < \epsilon\text{.}

L5.5.5: Full scalar comparison proof, with Cauchy-to-arbitrary-endpoint passage explicitly supplied in P18.1.

Proof.

We start with the first item. For every bb and c,c\text{,} such that a≤b≤c,a \leq b \leq c\text{,} we have −g(x)≤f(x)≤g(x),-g(x) \leq f(x) \leq g(x)\text{,} and so
∫bc−g≤∫bcf≤∫bcg.\begin{equation*} \int_b^c -g \leq \int_b^c f \leq \int_b^c g . \end{equation*}
In other words, ∣∫bcf∣≤∫bcg.\abs{\int_b^c f} \leq \int_b^c g\text{.}
Let ϵ>0\epsilon > 0 be given. Because of Proposition 5.5.3,
∫a∞g=∫abg+∫b∞g.\begin{equation*} \int_a^\infty g = \int_a^b g + \int_b^\infty g . \end{equation*}
As ∫abg\int_a^b g goes to ∫a∞g\int_a^\infty g as bb goes to infinity, ∫b∞g\int_b^\infty g goes to 0 as bb goes to infinity. Choose BB such that
∫B∞g<ϵ.\begin{equation*} \int_B^\infty g < \epsilon . \end{equation*}
As gg is nonnegative, if B≤b<c,B \leq b < c\text{,} then ∫bcg<ϵ\int_b^c g < \epsilon as well. Let {xn}n=1∞\{ x_n \}_{n=1}^\infty be a sequence going to infinity. Let MM be such that xn≥Bx_n \geq B for all n≥M.n \geq M\text{.} Take n,m≥M,n, m \geq M\text{,} with xn≤xm,x_n \leq x_m\text{,}
∣∫axmf−∫axnf∣=∣∫xnxmf∣≤∫xnxmg<ϵ.\begin{equation*} \abs{\int_a^{x_m} f - \int_a^{x_n} f} = \abs{\int_{x_n}^{x_m} f} \leq \int_{x_n}^{x_m} g < \epsilon . \end{equation*}
Therefore, the sequence {∫axnf}n=1∞\bigl\{ \int_a^{x_n} f \bigr\}_{n=1}^\infty is Cauchy and hence converges.
We need to show that the limit is unique. Suppose {xn}n=1∞\{ x_n \}_{n=1}^\infty is a sequence converging to infinity such that {∫axnf}n=1∞\bigl\{ \int_a^{x_n} f \bigr\}_{n=1}^\infty converges to L1,L_1\text{,} and {yn}n=1∞\{ y_n \}_{n=1}^\infty is a sequence converging to infinity such that {∫aynf}n=1∞\bigl\{ \int_a^{y_n} f \bigr\}_{n=1}^\infty converges to L2.L_2\text{.} Then there must be some nn such that ∣∫axnf−L1∣<ϵ\babs{\int_a^{x_n} f - L_1} < \epsilon and ∣∫aynf−L2∣<ϵ.\babs{\int_a^{y_n} f - L_2} < \epsilon\text{.} We can also suppose xn≥Bx_n \geq B and yn≥B.y_n \geq B\text{.} Then
∣L1−L2∣≤∣L1−∫axnf∣+∣∫axnf−∫aynf∣+∣∫aynf−L2∣<ϵ+∣∫xnynf∣+ϵ<3ϵ.\begin{equation*} \sabs{L_1 - L_2} \leq \abs{L_1 - \int_a^{x_n} f} + \abs{\int_a^{x_n} f- \int_a^{y_n} f} + \abs{\int_a^{y_n} f - L_2} < \epsilon + \abs{\int_{x_n}^{y_n} f} + \epsilon < 3 \epsilon. \end{equation*}
As ϵ>0\epsilon > 0 was arbitrary, L1=L2,L_1 = L_2\text{,} and hence ∫a∞f\int_a^\infty f converges. Above we have shown that ∣∫acf∣≤∫acg\abs{\int_a^c f} \leq \int_a^c g for all c>a.c > a\text{.} By taking the limit c→∞,c \to \infty\text{,} the first item is proved.
The second item is simply a contrapositive of the first item.

L5.5.2-gt1: The actually written p>1 infinite-right-endpoint case only; other p-test cases excluded.

Proof.

The proof follows by application of the fundamental theorem of calculus. Let us do the proof for p>1p > 1 for the infinite right endpoint and leave the rest to the reader. Hint: You should handle p=1p=1 separately.
Suppose p>1.p > 1\text{.} Then using the fundamental theorem,
∫1b1xp dx=∫1bx−p dx=b−p+1−p+1−1−p+1−p+1=−1(p−1)bp−1+1p−1.\begin{equation*} \int_1^b \frac{1}{x^p} \,dx = \int_1^b x^{-p} \,dx = \frac{b^{-p+1}}{-p+1} - \frac{1^{-p+1}}{-p+1} = \frac{-1}{(p-1)b^{p-1}} + \frac{1}{p-1} . \end{equation*}
As p>1,p > 1\text{,} we have p−1>0.p-1 > 0\text{.} Take the limit as b→∞b \to \infty to obtain that 1bp−1\frac{1}{b^{p-1}} goes to 0. The result follows.