Improper Integrals: Examples and Counterexamples

Substitution $x=-t$ gives [L1] ${\int }_{-1}^{-\epsilon }\frac{dx}{x}=-{\int }_{\epsilon }^1\frac{dt}{t}\mathrm{.}$ Thus the symmetric sum is exactly zero for every $\epsilon >0$, and [L1] gives principal value zero.

By [L2], the right-hand integral diverges to $+\mathrm{\infty }$; the identity in step 1.1 makes the left-hand one diverge to $-\mathrm{\infty }$. Hence separate improper convergence fails, showing that the converse of the principal-value theorem is false.

Every compact interval meets only finitely many jumps, so [L3] makes $f$ properly integrable there. At a positive integer $N$, [L1, L3] ${\int }_0^{N}f={\sum }_{k=0}^{N-1}\frac{(-1{)}^k}{k+1},$ which converges as $N\to \mathrm{\infty }$ by [L1].

If $N\le R<N+1$, the remaining integral from $N$ to $R$ has absolute value at most $1\mathrm{/}(N+1)$. Hence arbitrary real truncations have the same limit as the integer truncations.

At integer truncations, ${\int }_0^N\mathrm{\mid }f\mathrm{\mid }={\sum }_{k=0}^{N-1}1\mathrm{/}(k+1)$, unbounded by [L2]. Thus the integral converges but not absolutely.

By [L1], local finiteness and matching zero endpoint values make the spike sum continuous; adding the positive continuous baseline preserves positivity and continuity. Direct differentiation gives primitive $1-1\mathrm{/}(1+x)$ for the baseline, whose improper integral is one.

By [L2], additivity, and [L3], the total integral of all spikes is ${\sum }_{k=1}^{\mathrm{\infty }}{2}^{-k-2}<\mathrm{\infty }$. Given $\epsilon >0$, choose $K$ so that both the geometric spike tail from $K$ and the baseline tail $1\mathrm{/}(1+K)$ are below $\epsilon \mathrm{/}2$. For $K\le u<v$, nonnegativity bounds the integral over every partial spike by the full spike area, so ${\int }_u^{v}f<\epsilon$. The Cauchy criterion therefore gives convergence of ${\int }_0^{\mathrm{\infty }}f$.

At every positive integer $k$, $f(k)\ge s_k(k)=k$. Integers occur arbitrarily far out, so $f$ is unbounded on every tail.

Since $1\mathrm{/}2<1$, convergence follows from [L1]. For $0<c<1$, [L2] gives [L1, L2] ${\int }_c^1{x}^{-1\mathrm{/}2}dx=2(1-\sqrt{c})\mathrm{.}$ As $c\downarrow 0$, $\sqrt{c}\to 0$, so the limit is two.

This is not a proper Riemann integral on $[0,1]$: the displayed integrand is unbounded at zero, whereas every properly Riemann-integrable function is bounded.

Applying [L1] at $p=2$ proves convergence only at infinity, and [L2] gives the value $1\mathrm{/}(2-1)=1$ there.

Applying [L1] at $p=1\mathrm{/}2$ proves convergence only at zero, and [L2] gives $1\mathrm{/}(1-1\mathrm{/}2)=2$. These are precisely the four assertions displayed above.

Here $f$ is continuous and $f(0)=1$. Also $0<f(t)=1\mathrm{/}(1+t)<\epsilon$ whenever $t>1\mathrm{/}\epsilon$, so $f(t)\to 0$ directly from the definition of the limit at infinity. Thus [L1] gives exactly the displayed identity.

For $x>0$, the integrand simplifies to [L2] $\frac{b-a}{(1+ax)(1+bx)}\mathrm{.}$ It has finite limit $b-a$ at zero and is bounded in absolute value by a constant multiple of $x^{-2}$ for $x\ge 1$. This independently confirms local convergence at zero and tail convergence by [L2], without replacing the proper factor by a logarithm. ∎

By [L1], the integral over $[-1,1]$ is proper. On $[1,\mathrm{\infty })$, [L2] and [L3] prove convergence. Substitution $t=-x$ gives the identical conclusion on $(-\mathrm{\infty },-1]$.

Since $f\ge 0$, $\mathrm{\mid }f\mathrm{\mid }=f$. Both tails converge separately and the middle piece is proper, so the mixed integral is absolutely convergent by definition.

The map $\varphi$ is decreasing, so [L3] uses $\mathrm{\mid }{\varphi }^{\mathrm{\prime }}\mathrm{\mid }=t^{-2}$. By [L1] and [L2], the transformed integrand is $t^{p-2}$, proving the identity.

Write $t^{p-2}=t^{-(2-p)}$. The finite-endpoint $p$-test says this converges exactly when $2-p<1$, namely $p>1$, which is also precisely the infinite-endpoint threshold for the original integral.

For a positive integer $N$, additivity gives the displayed sum, whose parity values follow from the alternating sequence. [L1, L2] ${\int }_0^{N}f={\sum }_{k=0}^{N-1}(-1{)}^k,$ which equals one for odd $N$ and zero for even $N$ by [L2]. Thus the integer truncations are bounded but have no limit. [L1, L2] If $N\le R<N+1$, the remaining integral has absolute value at most one, so the full truncation primitive is bounded as asserted.

An improper limit would restrict to the same limit along all integer truncations, contradicting step 1.1. The example changes sign, so it does not satisfy the nonnegativity hypothesis in [L3] and shows that hypothesis cannot be deleted.

On the left use $t=c-x$; on the right use $t=x-c$. The two one-sided integrals become respectively ${\int }_0^{c-a}{t}^{-1\mathrm{/}2}dt$ and ${\int }_0^{b-c}{t}^{-1\mathrm{/}2}dt$.

Both converge separately by [L2], as [L1] requires. Evaluating them with [L3] and adding gives the displayed value.

For $0<\epsilon <1$, symmetry and [L2] give [L2] ${\int }_{-1}^{-\epsilon }\frac{dx}{x^2}+{\int }_{\epsilon }^1\frac{dx}{x^2}=2(\frac{1}{\epsilon }-1)\mathrm{.}$

This tends to $+\mathrm{\infty }$ as $\epsilon \downarrow 0$, not to a finite real. Therefore the principal value in [L1] diverges.

For $0<x\le 1$, one has $1\le 1+x\le 2$. If $q\ge 0$, [L1] gives $2^{-q}\le (1+x{)}^{-q}\le 1$; if $q<0$, it gives $1\le (1+x{)}^{-q}\le 2^{-q}$. Thus the kernel is bounded above and below by positive constant multiples of $x^{-p}$, so [L2] and [L3] give convergence at zero exactly when $p<1$.

For $x\ge 1$, $1\mathrm{/}2\le x\mathrm{/}(1+x)<1$. By [L1], the quotient of the kernel by $x^{-(p+q)}$ is $(x\mathrm{/}(1+x){)}^q$ and lies between $2^{-q}$ and $1$ when $q\ge 0$, and between $1$ and $2^{-q}$ when $q<0$. Hence [L2] and [L3] give convergence at infinity exactly when $p+q>1$.

The mixed definition requires both ends separately, so the full integral converges exactly under the two simultaneous inequalities. Positive bases ensure every rational power used above is defined, regardless of the signs of $p$ and $q$.