Why trigonometric integrals can be improper
A trigonometric function may be smooth across most of a finite interval while becoming unbounded at particular points. For example, \(\sec x=1/\cos x\) is undefined wherever \(\cos x=0\), and \(\csc x=1/\sin x\) is undefined wherever \(\sin x=0\). These points divide an integral into pieces, and convergence must be checked on each piece separately.
The key observation is that sine and cosine approach zero linearly near each of their zeros. Consequently, a reciprocal power of sine or cosine behaves locally like a reciprocal power of the distance to the singular point. The power-singularity threshold then determines convergence.
Locating the singular points
The zeros of cosine occur at \(x=\frac{\pi}{2}+k\pi\), and the zeros of sine occur at \(x=k\pi\), where \(k\in\mathbb{Z}\). Thus \(\sec x\) has singularities at the cosine zeros, while \(\csc x\) has singularities at the sine zeros. On any finite interval there are only finitely many such points.
Before testing convergence, list every singular point that lies in the interval. Interior singularities become split points. If a singular point is an endpoint, the integral is instead evaluated using a one-sided limit there. A singularity outside the interval is irrelevant to that integral.
Why the blow-up is a power of the distance
Consider a zero \(a\) of sine or cosine, and write \(x=a+t\), so \(t\) is the distance in signed form from \(a\). The standard limit \(\lim_{t\to 0}\frac{\sin t}{t}=1\) implies that, for all sufficiently small nonzero \(t\), \(\frac12\leq\left|\frac{\sin t}{t}\right|\leq\frac32\). Equivalently, \(\frac12|t|\leq|\sin t|\leq\frac32|t|\).
At a cosine zero \(a=\frac{\pi}{2}+k\pi\), the identity \(\cos(a+t)=(-1)^{k+1}\sin t\) shows that \(|\cos(a+t)|=|\sin t|\). At a sine zero \(a=k\pi\), the identity \(\sin(a+t)=(-1)^k\sin t\) gives \(|\sin(a+t)|=|\sin t|\). Therefore, near either type of zero, the absolute value of the vanishing denominator is bounded above and below by positive constant multiples of \(|t|\).
For any fixed \(q>0\), taking reciprocal \(q\)-th powers reverses the bounds. Thus \(|\sec(a+t)|^q\), near a cosine zero, and \(|\csc(a+t)|^q\), near a sine zero, are each bounded above and below by positive constant multiples of \(|t|^{-q}\). The earlier power-singularity threshold states that \(\int_0^\delta t^{-q}\,dt\) converges exactly when \(q<1\), for any \(\delta>0\).
The comparison works on each side because the bounds hold for both positive and negative \(t\). Away from the finitely many denominator zeros, the trigonometric integrand is continuous on closed subintervals and hence bounded there. Local behavior at the zeros is therefore what decides convergence.
Worked Example: A square-root secant blow-up
Worked Example: A square-root secant blow-up
Determine whether \(\int_0^\pi|\sec x|^{1/2}\,dx\) converges.
The denominator \(\cos x\) vanishes in \([0,\pi]\) only at \(a=\frac{\pi}{2}\). Thus the integral must be split there. Let \(t=x-a\). Since \(|\cos(a+t)|=|\sin t|\), the local bounds give \(\frac12|t|\leq|\cos(a+t)|\leq\frac32|t|\). Taking inverse square roots yields \((\frac32)^{-1/2}|t|^{-1/2}\leq|\sec(a+t)|^{1/2}\leq(\frac12)^{-1/2}|t|^{-1/2}\).
On each side of \(a\), the comparison integral is finite: \(\int_0^\delta t^{-1/2}\,dt=2\sqrt{\delta}\). So the one-sided integral on the left converges, and the one-sided integral on the right converges as well. On the rest of \([0,\pi]\), away from \(a\), the integrand is continuous and bounded. Therefore the full improper integral converges.
Check: The only denominator zero is \(\frac{\pi}{2}\); both one-sided pieces have exponent \(q=\frac12<1\), and there are no other singularities. This verifies that every part of the interval has been accounted for.
Worked Example: Secant across an interior pole
Worked Example: Secant across an interior pole
Determine whether \(\int_0^\pi\sec x\,dx\) converges as an ordinary improper integral.
The denominator vanishes at \(a=\frac{\pi}{2}\), so the required pieces are \(\int_0^a\sec x\,dx\) and \(\int_a^\pi\sec x\,dx\), each defined by a one-sided limit. An antiderivative on intervals where \(\cos x\neq0\) is \(F(x)=\ln|\sec x+\tan x|\). Indeed, \(\frac{d}{dx}F(x)=\frac{\sec x\tan x+\sec^2x}{\sec x+\tan x}=\sec x\), wherever this expression is defined.
On the left, the improper integral is \(\lim_{\varepsilon\to0^+}\bigl(F(\frac{\pi}{2}-\varepsilon)-F(0)\bigr)\). Here \(F(0)=0\), while \(\sec x+\tan x\) tends to \(+\infty\) as \(x\to\frac{\pi}{2}^{-}\). Thus \(F(\frac{\pi}{2}-\varepsilon)\to+\infty\), and the left piece diverges.
On the right, the improper integral is \(\lim_{\varepsilon\to0^+}\bigl(F(\pi)-F(\frac{\pi}{2}+\varepsilon)\bigr)\). Since \(F(\pi)=0\), while \(|\sec x+\tan x|\to\infty\) as \(x\to\frac{\pi}{2}^{+}\), we have \(F(\frac{\pi}{2}+\varepsilon)\to+\infty\). The right piece therefore tends to \(-\infty\) and diverges. Since neither one-sided integral converges to a finite value, the ordinary improper integral diverges.
Check: Differentiating \(F\) gives \(\sec x\), and the limits on the two split intervals diverge separately. The conclusion follows from the definition of an interior singularity, not from combining the two limits.
Worked Example: Several cosecant singularities
Worked Example: Several cosecant singularities
Determine whether \(\int_0^{2\pi}\csc^2 x\,dx\) converges.
The denominator \(\sin x\) vanishes at \(0,\pi,2\pi\). The points \(0\) and \(2\pi\) are singular endpoints, and \(\pi\) is an interior singularity. An antiderivative wherever \(\sin x\neq0\) is \(-\cot x\), since \(\frac{d}{dx}(-\cot x)=\csc^2 x\).
At the left endpoint, the one-sided integral from \(0\) to \(\frac{\pi}{2}\) has limit \(\lim_{\varepsilon\to0^+}\int_\varepsilon^{\pi/2}\csc^2x\,dx =\lim_{\varepsilon\to0^+}\cot\varepsilon=+\infty\). At the left side of the interior singularity, \(\lim_{\varepsilon\to0^+}\int_{\pi/2}^{\pi-\varepsilon}\csc^2x\,dx =\lim_{\varepsilon\to0^+}\cot\varepsilon=+\infty\). At its right side, \(\lim_{\varepsilon\to0^+}\int_{\pi+\varepsilon}^{3\pi/2}\csc^2x\,dx =\lim_{\varepsilon\to0^+}\cot\varepsilon=+\infty\). Finally, approaching the right endpoint from the left gives \(\lim_{\varepsilon\to0^+}\int_{3\pi/2}^{2\pi-\varepsilon}\csc^2x\,dx =\lim_{\varepsilon\to0^+}\cot\varepsilon=+\infty\).
These limits use \(\cot\varepsilon\to+\infty\) as \(\varepsilon\to0^+\), and the displayed evaluations follow from the antiderivative \(-\cot x\). Each one-sided piece diverges, so the full improper integral diverges.
Check: The zero at \(\pi\) requires two separate one-sided tests, while \(0\) and \(2\pi\) each require an endpoint test. The exponent is \(q=2\), above the convergence threshold \(q<1\), consistent with every computed limit.
A common mistake: evaluating across a pole
A tempting but invalid calculation is to apply the secant antiderivative from the preceding example directly across the singularity: \(\int_0^\pi\sec x\,dx=F(\pi)-F(0)=0-0=0\). This treats the interval as if the integrand were continuous throughout it. It is not: \(\sec x\) is undefined and unbounded at \(\frac{\pi}{2}\). The fundamental theorem of calculus cannot evaluate an integral across that pole by subtracting endpoint values.
The correction is to split at \(\frac{\pi}{2}\), form the two one-sided limits, and test them separately. As shown above, they diverge in opposite signed directions, so the ordinary improper integral does not converge. An apparent cancellation between divergent pieces does not make either one-sided integral finite.
Putting the test into practice
For secant, solve \(\cos x=0\); for cosecant, solve \(\sin x=0\). Keep only points in the interval, including its endpoints.
Use every interior singularity as a split point. Treat a singular endpoint with a one-sided limit.
Near a denominator zero, reciprocal power \(q\) behaves like \(|x-a|^{-q}\). Test the corresponding one-sided integrals using the threshold \(q<1\).
The ordinary improper integral converges only if all required one-sided integrals converge. Do not combine divergent limits across a singularity.
This approach matters because a trigonometric integrand can look familiar while its finite-interval integral is undefined as an ordinary improper integral. Locating the zeros first prevents missed poles, and the local power test turns the convergence question into a precise comparison near each one.
Check Your Understanding
For each question, identify the relevant singular points and explain what must be checked.
- Where are the singularities of \(\sec x\) on \([0,2\pi]\)? Which are interior and which are endpoints?
- For \(q>0\), when does \(|\csc x|^q\) have a convergent one-sided integral near a zero of sine?
- Why must an integral with an interior trigonometric pole be split into two one-sided integrals?
- Does \(\int_0^\pi|\sec x|^{3/4}\,dx\) converge? State the local test that decides.
- Why is it invalid to evaluate an antiderivative at the two endpoints when a pole lies between them?
