Absolute And Conditional Convergence

Converges absolutely.

We have $\left|{\displaystyle \frac{\sin x}{x^{3/2}}}\right|\le {\displaystyle \frac{1}{x^{3/2}}}$ and ${\displaystyle {\int }_1^{\mathrm{\infty }}{x}^{-3/2}dx}$ converges (the $p$-test with ). By the Direct Comparison Test, ${\displaystyle {\int }_1^{\mathrm{\infty }}\left|\frac{\sin x}{x^{3/2}}\right|dx}$ converges. Since absolute convergence implies convergence, the integral converges, and it converges absolutely.

Converges conditionally.

(i) Convergence. We use integration by parts with $u=1/\sqrt{x}$ and $dv=\sin xdx$, giving $du=-1/(2x^{3/2})dx$ and $v=-\cos x$. Then:

As $b\to \mathrm{\infty }$: $-\cos b/\sqrt{b}\to 0$ (since $|\cos b/\sqrt{b}|\le 1/\sqrt{b}\to 0$), and the remaining integral converges absolutely since $|\cos x/(2x^{3/2})|\le 1/(2x^{3/2})$ and ${\displaystyle {\int }_1^{\mathrm{\infty }}{x}^{-3/2}dx}$ converges. Therefore the integral converges.

(ii) Non-absolute convergence. On $[n\pi ,(n+1)\pi ]$ we have $1/\sqrt{x}\ge 1/\sqrt{(n+1)\pi }$, so:

Since ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}1/\sqrt{n+1}}$ diverges (comparison with the harmonic series via $1/\sqrt{n+1}\ge 1/(n+1)$, or more directly since $\sum n^{-1/2}$ diverges), we conclude that ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{|\sin x|}{\sqrt{x}}dx}$ diverges. Therefore the convergence is conditional.

${\displaystyle \frac{1}{2}}$.

Since $|e^{-x}\sin x|\le e^{-x}$ and ${\displaystyle {\int }_0^{\mathrm{\infty }}{e}^{-x}dx=1}$ converges, the Direct Comparison Test gives convergence of ${\displaystyle {\int }_0^{\mathrm{\infty }}{e}^{-x}|\sin x|dx}$. Therefore the integral converges absolutely.

To evaluate the integral, we integrate by parts twice. Let $I={\displaystyle {\int }_0^{\mathrm{\infty }}{e}^{-x}\sin xdx}$.

First integration by parts. Let $u=\sin x$ and $dv=e^{-x}dx$, so $du=\cos xdx$ and $v=-e^{-x}$:

Second integration by parts. Let $u=\cos x$ and $dv=e^{-x}dx$, so $du=-\sin xdx$ and $v=-e^{-x}$:

Solving for $I$. Substituting back:

Adding ${\displaystyle {\int }_0^b{e}^{-x}\sin xdx}$ to both sides:

Since $|e^{-b}(\sin b+\cos b)|\le 2e^{-b}\to 0$ as $b\to \mathrm{\infty }$, taking the limit gives $2I=1$, so $I={\displaystyle \frac{1}{2}}$.

Converges conditionally.

(i) Convergence. We use integration by parts with $u=1/x$ and $dv=\cos (\pi x)dx$, so $du=-1/x^{2}dx$ and $v=\sin (\pi x)/\pi$. Then:

As $b\to \mathrm{\infty }$: $\sin (\pi b)/(\pi b)\to 0$, and $\sin (\pi \cdot 1)/(\pi \cdot 1)=0$ since $\sin \pi =0$, so the boundary term tends to $0$. For the remaining integral, $|\sin (\pi x)/(\pi x^2)|\le 1/(\pi x^2)$ and ${\displaystyle {\int }_1^{\mathrm{\infty }}dx/x^2}$ converges, so by the Direct Comparison Test the remaining integral converges absolutely. Therefore ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{\cos (\pi x)}{x}dx}$ converges.

(ii) Non-absolute convergence. The function $\cos (\pi x)$ has period $2$, so on each interval $[n,n+1]$ with $n\ge 1$ it completes exactly half a period and $|\cos (\pi x)|$ traces one full arch. Hence ${\displaystyle {\int }_n^{n+1}|\cos (\pi x)|dx=\frac{2}{\pi }}$, the same value for every $n$. Also $1/x\ge 1/(n+1)$ on $[n,n+1]$. Therefore:

Summing over $n=1,\dots ,N$:

Since the harmonic series diverges, ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{|\cos (\pi x)|}{x}dx}$ diverges. The convergence of ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{\cos (\pi x)}{x}dx}$ is therefore conditional.

Converges absolutely, and its value is at most $\pi /4\approx 0.7854$ in absolute value.

The integrand changes sign, so we test $|f|$ first. Since $|\cos x|\le 1$ and for $x\ge 1$,

Since ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{dx}{x^2}=1}$ converges (the $p$-test with ), the Direct Comparison Test gives convergence of ${\displaystyle {\int }_1^{\mathrm{\infty }}\left|\frac{\cos x}{x^2+1}\right|dx}$. The integral therefore converges absolutely, and by the theorem of this section it converges.

For the bound, use the triangle inequality together with the sharper comparison function $1/(x^2+1)$:

This is a typical dividend of absolute convergence: the same comparison that proves convergence also produces an error bound.

${\displaystyle \frac{1}{2}}$.

Absolute convergence. Since $|\cos x|\le 1$ we have $|e^{-x}\cos x|\le e^{-x}$ for $x\ge 0$, and

converges. By the Direct Comparison Test, ${\displaystyle {\int }_0^{\mathrm{\infty }}{e}^{-x}|\cos x|dx}$ converges, so the integral converges absolutely.

Evaluation. Let $J={\displaystyle {\int }_0^{\mathrm{\infty }}{e}^{-x}\cos xdx}$. Integrate by parts with $u=\cos x$, $dv=e^{-x}dx$, so $du=-\sin xdx$ and $v=-e^{-x}$:

Now integrate the remaining integral by parts with $u=\sin x$, $dv=e^{-x}dx$, so $du=\cos xdx$ and $v=-e^{-x}$:

Substituting the second computation into the first:

Moving the last integral to the left side gives

Since $|e^{-b}(\cos b-\sin b)|\le 2e^{-b}\to 0$ as $b\to \mathrm{\infty }$, we get $2J=1$, so $J={\displaystyle \frac{1}{2}}$.

Note that this is the same value as ${\displaystyle {\int }_0^{\mathrm{\infty }}{e}^{-x}\sin xdx}$ from an earlier exercise; the two integrals are equal, although the functions are not.

Converges conditionally. It is surprising because $\sin (x^2)$ does not tend to $0$ as $x\to \mathrm{\infty }$.

Substitution. On any finite interval $[1,b]$, put $u=x^2$. Then $du=2xdx$, and since we have $dx={\displaystyle \frac{du}{2\sqrt{u}}}$. The limits become $u=1$ and $u=b^2$, so

As $b\to \mathrm{\infty }$ we also have $b^2\to \mathrm{\infty }$, so the left side has a limit exactly when the right side does:

Convergence. The integral ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{\sin u}{\sqrt{u}}du}$ was shown to converge in the exercise on sine over the square root, by integration by parts. Hence ${\displaystyle {\int }_1^{\mathrm{\infty }}\sin (x^2)dx}$ converges.

Not absolutely. The same substitution applied to $|\sin (x^2)|$ gives

and part (ii) of that same exercise shows the right side is $+\mathrm{\infty }$. So the convergence is conditional.

Why it is surprising. For an infinite series, convergence of $\sum a_n$ forces $a_n\to 0$. Here $\sin (x^2)$ returns to $1$ whenever $x=\sqrt{\pi /2+2k\pi }$ and to $-1$ whenever $x=\sqrt{3\pi /2+2k\pi }$, so it never settles down. What does happen is that the successive zeros of $\sin (x^2)$, namely $x=\sqrt{n\pi }$, crowd together as $x$ grows: consecutive zeros are about ${\displaystyle \frac{\pi }{2\sqrt{n\pi }}}$ apart. The lobes become narrow fast enough that their areas shrink, and the alternating signs finish the job. Numerically the integral equals about $0.3164$.

Converges conditionally.

Reduce to a series. For we have $\lfloor x\rfloor =n$, so the integrand equals $(-1{)}^n/x$ on that interval. The integrand has a jump at each integer but is bounded, so it is integrable on every bounded subinterval, and

Adding these for $n=1,\dots ,N-1$ gives

Convergence. The terms $a_n=\ln (1+\frac{1}{n})$ are positive, strictly decreasing (because $1+\frac{1}{n}$ decreases), and tend to $\ln 1=0$. By the Alternating Series Test the series converges, so the integrals over integer upper limits converge to some number $L$. For a general upper limit $b$ with ,

so ${\displaystyle {\int }_1^b}$ has the same limit $L$ as $b\to \mathrm{\infty }$ through all real values. The integral converges.

Not absolutely. Since $|(-1{)}^{\lfloor x\rfloor }|=1$ for every $x$,

which diverges by the $p$-test with $p=1$. Therefore the convergence is conditional.

This integral is the exact analogue of the alternating harmonic series: it converges for precisely the same reason, and it fails to converge absolutely for precisely the same reason. Its value can be evaluated with Wallis's product and equals $\ln (2/\pi )\approx -0.4516$.

Diverges, because of the singularity at $x=0$. The tail ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{\sin x}{x^2}dx}$ by itself converges absolutely.

Step 1: split at the two improper points. The integral is improper at $x=0$, where the integrand is unbounded, and at $+\mathrm{\infty }$. Split at $x=1$:

Step 2: the tail is harmless. For $x\ge 1$ we have $\left|{\displaystyle \frac{\sin x}{x^2}}\right|\le {\displaystyle \frac{1}{x^2}}$, and ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{dx}{x^2}=1}$ converges. By the Direct Comparison Test the tail converges absolutely.

Step 3: the origin is not. Near $x=0$ we have $\sin x\approx x$, so the integrand behaves like $x/x^2=1/x$. Confirm this with the Limit Comparison Test using $g(x)=1/x$, which is legitimate because on $(0,1]$, so the integrand is positive there:

Since $L=1$ lies in $(0,\mathrm{\infty })$ and ${\displaystyle {\int }_0^1\frac{dx}{x}}$ diverges (the $p$-test for Type II integrals with $p=1$), the piece ${\displaystyle {\int }_0^1\frac{\sin x}{x^2}dx}$ diverges.

Conclusion. One divergent piece makes the whole integral divergent, so ${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{\sin x}{x^2}dx}$ diverges.

Compare this with ${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}dx}$, where the extra factor of $x$ in the denominator is absent and the singularity at the origin is removable, since $(\sin x)/x\to 1$. Raising the exponent from $1$ to $2$ improves the behaviour at infinity and ruins it at the origin.

False. The conclusion does hold if, in addition, $f$ is monotone on $[a,\mathrm{\infty })$, and also if $f$ is uniformly continuous on $[a,\mathrm{\infty })$.

A first counterexample. By the exercise on ${\int }_1^{\mathrm{\infty }}\sin (x^2)dx$, that integral converges, yet $\sin (x^2)$ takes the value $1$ at $x=\sqrt{\pi /2+2k\pi }$ and the value $-1$ at $x=\sqrt{3\pi /2+2k\pi }$ for every $k$. So $f(x)=\sin (x^2)$ has no limit at all as $x\to \mathrm{\infty }$, let alone the limit $0$.

A second counterexample, with spikes. Let $f$ be the function that equals $1$ on each interval $[n,n+2^{-n}]$ for $n=1,2,3,\dots$ and equals $0$ elsewhere. (These intervals are disjoint, since .) On any bounded interval $f$ has finitely many discontinuities, so it is integrable there, and for every $b$,

The function $b\mapsto {\int }_1^{b}f$ is nondecreasing and bounded above, so it has a limit; the integral converges (to $1$). But $f(n)=1$ for every integer $n$, so $f(x)\nrightarrow 0$. Replacing the rectangles by narrow continuous triangles of height $1$ and base $2^{-n}$ makes the same example continuous.

When the conclusion is true. If $f$ is monotone on $[a,\mathrm{\infty })$ and ${\displaystyle {\int }_a^{\mathrm{\infty }}f}$ converges, then $f(x)\to 0$: a monotone function has a limit $L$ at infinity, and $L\ne 0$ would force the integral of a function eventually bounded away from $0$ to diverge. The same conclusion holds if $f$ is uniformly continuous on $[a,\mathrm{\infty })$: uniform continuity stops the spikes from becoming arbitrarily narrow, since if $|f(x_k)|\ge \epsilon$ for some sequence $x_k\to \mathrm{\infty }$, then $|f|\ge \epsilon /2$ on an interval of one fixed length around each $x_k$, and those intervals contribute the same positive amount of area infinitely often.

The moral is that convergence of an improper integral is a statement about accumulated area, not about the size of the integrand. This is the main way integrals differ from series.

The product integral converges absolutely, with $\left|{\displaystyle {\int }_a^{\mathrm{\infty }}fg}\right|\le M{\displaystyle {\int }_a^{\mathrm{\infty }}|f|}$. The statement fails for conditionally convergent $f$: take $f(x)=(\sin x)/x$ and $g(x)=\sin x$ on $[1,\mathrm{\infty })$.

Proof. The product $fg$ is integrable on every bounded subinterval, being a product of two such functions. For every $x\ge a$,

and $M|f|$ is nonnegative with ${\displaystyle {\int }_a^{\mathrm{\infty }}M|f(x)|dx=M{\int }_a^{\mathrm{\infty }}|f(x)|dx}$ convergent by hypothesis. Since $|fg|\ge 0$, the Direct Comparison Test applies and gives convergence of ${\displaystyle {\int }_a^{\mathrm{\infty }}|f(x)g(x)|dx}$. By the theorem of this section, ${\displaystyle {\int }_a^{\mathrm{\infty }}f(x)g(x)dx}$ converges, and it converges absolutely. Taking absolute values through the limit gives the bound

Failure for conditional convergence. Let $a=1$, $f(x)={\displaystyle \frac{\sin x}{x}}$ and $g(x)=\sin x$, so $|g|\le 1$ is bounded and ${\displaystyle {\int }_1^{\mathrm{\infty }}f}$ converges conditionally. Their product is

using the identity ${\sin }^{2}x=\frac{1}{2}(1-\cos 2x)$. Now ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{\cos 2x}{2x}dx}$ converges (integrate by parts exactly as for $(\cos x)/x$), while ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{dx}{2x}}$ diverges. A divergent integral minus a convergent one is divergent, so ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{{\sin }^{2}x}{x}dx}$ diverges.

Multiplying by a bounded function therefore destroyed the convergence. This is the precise sense in which conditional convergence is fragile: it depends on cancellation, and a bounded factor can remove the cancellation.

(a) converges conditionally. (b) converges absolutely. (c) diverges.

(a) First test $|f|$. On $[n\pi ,(n+1)\pi ]$ the graph of $\cos x$ runs from $\pm 1$ down through a zero at the midpoint to $\mp 1$, so $|\cos x|$ traces two half-arches whose areas together give ${\displaystyle {\int }_{n\pi }^{(n+1)\pi }|\cos x|dx=2}$, and $x^{-1/3}\ge ((n+1)\pi {)}^{-1/3}$ there. Hence

and ${\displaystyle \sum _n(n+1{)}^{-1/3}}$ diverges (its terms exceed those of the harmonic series), so ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{|\cos x|}{x^{1/3}}dx}$ diverges. Absolute convergence fails.

Now test ${\displaystyle {\int }_1^{\mathrm{\infty }}f}$ itself by parts, with $u=x^{-1/3}$ and $dv=\cos xdx$, so $du=-\frac{1}{3}{x}^{-4/3}dx$ and $v=\sin x$:

The boundary term tends to $-\sin 1$, since $|\sin b|/b^{1/3}\le b^{-1/3}\to 0$. The remaining integral converges absolutely because $|\sin x/x^{4/3}|\le x^{-4/3}$ and . So the integral converges, and since $\int |f|$ diverges the convergence is conditional.

(b) Since for $x\ge 1$ and $|\sin x|\le 1$,

and ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{dx}{x^2}}$ converges. By the Direct Comparison Test the integral converges absolutely. (This is also the previous exercise applied with $f(x)=\sin x/x^2$, which is absolutely convergent, and the bounded factor $g(x)=\arctan x$.)

(c) Here the oscillation grows instead of decaying. An antiderivative of $x\sin x$ is $\sin x-x\cos x$, as differentiating confirms:

Therefore

Taking $b=2N\pi$ gives $-2N\pi \to -\mathrm{\infty }$ for the first bracket, while $b=(2N+1)\pi$ gives $+(2N+1)\pi \to +\mathrm{\infty }$. The partial integrals oscillate with unbounded amplitude, so no limit exists and the integral diverges.

Absolutely convergent for ; conditionally convergent for ; divergent for $p\le 0$.

Case 1: , absolute convergence. Since $|\sin x|\le 1$,

and ${\displaystyle {\int }_1^{\mathrm{\infty }}{x}^{-p}dx}$ converges by the $p$-test because . The Direct Comparison Test gives convergence of ${\displaystyle {\int }_1^{\mathrm{\infty }}\left|\frac{\sin x}{x^p}\right|dx}$, so the integral converges absolutely.

Case 2: , conditional convergence. Two things must be shown.

Convergence. Integrate by parts with $u=x^{-p}$ and $dv=\sin xdx$, so $du=-p{x}^{-p-1}dx$ and $v=-\cos x$:

Because , the boundary term satisfies $\left|{\displaystyle \frac{\cos b}{b^p}}\right|\le {\displaystyle \frac{1}{b^p}}\to 0$, so it tends to $\cos 1$. For the remaining integral, $\left|{\displaystyle \frac{\cos x}{x^{p+1}}}\right|\le {\displaystyle \frac{1}{x^{p+1}}}$ with , so ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{\cos x}{x^{p+1}}dx}$ converges absolutely. Both parts have limits, so the integral converges. This is the payoff of integration by parts: it raises the exponent from $p$ to $p+1$, moving it across the threshold $1$.

Non-absolute convergence. On $[n\pi ,(n+1)\pi ]$ we have $x^{-p}\ge ((n+1)\pi {)}^{-p}$, so

Summing over $n=1,\dots ,N$ bounds ${\displaystyle {\int }_{\pi }^{(N+1)\pi }\frac{|\sin x|}{x^p}dx}$ below by ${\displaystyle \frac{2}{{\pi }^p}}{\displaystyle \sum _{n=1}^N\frac{1}{(n+1{)}^p}}$, and the series $\sum (n+1{)}^{-p}$ diverges for $p\le 1$. Hence ${\displaystyle {\int }_1^{\mathrm{\infty }}\left|\frac{\sin x}{x^p}\right|dx}$ diverges and the convergence is conditional.

Case 3: $p\le 0$, divergence. Write $q=-p\ge 0$, so the integrand is $x^q\sin x$. On $[2n\pi ,(2n+1)\pi ]$ we have $\sin x\ge 0$ and $x^q\ge 1$ (since $x\ge 1$ and $q\ge 0$), so

for every $n\ge 1$. Now suppose the integral converged, say ${\displaystyle {\int }_1^b\frac{\sin x}{x^p}dx\to L}$ as $b\to \mathrm{\infty }$. Then the difference of the values at $b=(2n+1)\pi$ and $b=2n\pi$ would tend to $L-L=0$. But that difference is exactly the integral above, which is at least $2$ for every $n$. This contradiction shows the integral diverges.

Summary of the trichotomy. Reading the three cases together: the exponent $p=1$ separates absolute from conditional convergence, and the exponent $p=0$ separates conditional convergence from divergence. The case $p=1$ is the tail of the Dirichlet integral of this section, and $p=1/2$ is the exercise on sine over the square root. Every classification question about ${\int }_1^{\mathrm{\infty }}(\sin x)/x^{p}dx$ reduces to locating $p$ relative to these two thresholds.