Question bank — Improper integrals — Type I (infinite limits), Type II (discontinuous integrand)

False — width is infinite but height can shrink fast enough that the total area is finite; e.g. ${\int }_1^{\infty }{x}^{-2}dx=1$. Convergence is a race between growing width and shrinking height.

False — decay to zero is necessary-ish but nowhere near sufficient; $1/x\to 0$ yet ${\int }_1^{\infty }\frac{dx}{x}$ diverges because it decays too slowly ($p=1$, the knife-edge).

False — the exponent $p=0.999<1$, so it diverges. Being a hair below the boundary $p=1$ is still on the "diverges at infinity" side.

True — near zero the rule flips, and $p=0.999<1$ is exactly the condition for convergence, giving a finite $\frac{1}{1-p}=1000$.

False — area under a positive curve can only be positive (or $+\infty$); a negative "answer" is a red flag that you applied the Fundamental Theorem of Calculus illegally across a blow-up.

True — once each piece is a finite number, their sum is finite; and for a proper splitting point the total is independent of which legal $c$ you pick. The subtlety is only that you must actually place $c$ so that each piece isolates its own bad endpoint.

False — each half ${\int }_0^{\infty }xdx$ diverges, so the two-sided integral diverges; the "$0$" is only the Cauchy principal value, not convergence.

True — if $0\le g\le f$ and the smaller $\int g$ diverges, then $\int f$ diverges too; comparison works in both directions depending on which side you bound.

True — even though $\sin x$ oscillates in sign, $\left|\frac{\sin x}{x^2}\right|\le \frac{1}{x^2}$, and ${\int }_1^{\infty }{x}^{-2}$ converges; absolute convergence forces convergence.

True — but only conditionally: it converges by Dirichlet's test (bounded oscillating $\sin$ against the monotone-to-zero factor $1/x$), while ${\int }_1^{\infty }\frac{|\sin x|}{x}dx$ diverges. So it converges without converging absolutely.

The integrand blows up at $x=0$, which is inside $[-1,1]$, so the Fundamental Theorem of Calculus does not apply across it. You must split at $0$ and take limits — the right half alone is ${\lim }_{t\to 0^{+}}{\int }_t^1{x}^{-2}dx={\lim }_{t\to 0^{+}}\left(\frac{1}{t}-1\right)=+\infty$, so it diverges; the true value is $+\infty$, and the negative answer is impossible for a positive function.

Because you cannot substitute $\infty$ into $-e^{-x}$ as if it were a number; you must write ${\lim }_{t\to \infty }\left(-e^{-t}+1\right)$ and observe $e^{-t}\to 0$. The habit of plugging in infinity is exactly what makes people mishandle divergent cases.

Having an antiderivative is not enough — you must evaluate its limit: ${\lim }_{t\to \infty }\ln t=\infty$, so it diverges. The unbounded growth of $\ln t$ is exactly the divergence, and this is precisely why writing "$\ln \infty$" as a value is meaningless.

The comparison direction is useless: bounding your function below by a convergent one proves nothing. You need $\frac{1}{x^3+1}\le \frac{1}{x^3}$ (bound above by a convergent one) to conclude convergence.

A finite interval does not make it proper — the integrand blows up at the endpoint $0$, making it Type II; the limit ${\lim }_{t\to 0^{+}}(-\ln t)=+\infty$ diverges.

No — that symmetric limit is the principal value; the true definition splits at some $c$ and requires each one-sided piece to converge independently.

Because the Fundamental Theorem of Calculus only works on a closed, bounded interval with a continuous integrand; $t$ keeps us on legal ground, and the limit then asks whether the area settles on a number.

Because $\int x^{-1}dx=\ln x$, and $\ln$ grows/shrinks without bound too slowly-yet-surely at both ends: $\ln t\to \infty$ as $t\to \infty$ and $\ln t\to -\infty$ as $t\to 0^{+}$. It is the knife-edge where the power rule breaks.

Different enemies: at infinity you fight slow decay, so you want a big exponent to kill the tail; near zero you fight a tall spike, so you want a small exponent to keep the spike gentle.

Because "convergence" means the total area is a well-defined finite number regardless of how the two limits approach their bad points; if one half is $+\infty$, no legitimate cancellation with a separate $-\infty$ half is allowed.

The logic "smaller area under bigger finite area is finite" only holds when there's no sign cancellation; with mixed signs a smaller-looking function could still oscillate and behave differently, so you compare magnitudes or use an oscillation test like Dirichlet's instead (see Comparison Test for Integrals).

Because convergence here comes from cancellation: consecutive positive and negative humps nearly cancel, and their leftover shrinks like $1/x$. Dirichlet's test captures this — a bounded oscillation times a monotone factor decreasing to $0$ converges — a mechanism comparison, which ignores sign, simply cannot see.

That finite area is exactly what makes $e^{-x}$ a valid probability density on $[0,\infty )$ (it integrates to $1$), and the same convergence powers the Laplace Transform and Gamma Function.

It diverges at the top. Substitute $u=\ln x$, so $du=\frac{dx}{x}$, turning the integral into $\int \frac{du}{u}$ which runs to $\infty$ — the $p=1$ divergent integral in disguise; the extra $\ln x$ is too weak to save it. (It is also improper at $x=1$ where $\ln x\to 0$, so a careful treatment starts the integral at some $x=c>1$.)

It diverges for every $p$ — no $p$ can satisfy both "$p>1$ at infinity" and "$p<1$ at zero" simultaneously, so one end always kills it. You must split at $1$ and analyze each piece.

Yes — e.g. ${\int }_0^{\infty }\frac{dx}{\sqrt{x}(1+x)}$ has an infinite upper limit (Type I) and a blow-up at $0$ (Type II); split at a convenient point like $1$ and check each end separately.

Yes and it's the punchline — $1/x$ diverges at both ends, near-zero and near-infinity, which is why $p=1$ is called the lonely loser: it fails on every front. See p-series and Integral Test for the discrete mirror image.

Yes — with $u=\ln x$, $du=\frac{dx}{x}$, and the lower limit becomes $u=\ln e=1$, so it turns into ${\int }_1^{\infty }{u}^{-2}du=1$. The squared log is just enough to converge, where a single log failed; we deliberately start at $e$ so that $u\ge 1$ and the only issue is the infinite upper limit.

Because starting at $x=1$ makes $\ln x\to 0^{+}$, so $\frac{1}{x(\ln x{)}^2}\to +\infty$ there — the integral is also Type II at the lower end. The same substitution $u=\ln x$, $du=\frac{dx}{x}$ turns it into ${\int }_0^{\infty }{u}^{-2}du$, whose piece ${\int }_0^1{u}^{-2}du$ diverges. You must isolate and check the singularity at $x=1$ before celebrating the tail.

Not necessarily. Concretely, let $f$ be $0$ everywhere except a triangular spike of height $1$ and base-width $2^{-n}$ centred at each integer $x=n$; each spike has area $\frac{1}{2}\cdot 2^{-n}$, so the total area is $\sum \frac{1}{2}{2}^{-n}<\infty$ (converges), yet $f(n)=1$ for every $n$, so $f$ does not tend to $0$. For monotone or otherwise "nice" $f$, however, convergence does force $f\to 0$.

Substitute $u=\ln x$, $du=\frac{dx}{x}$: it becomes ${\int }_0^{\infty }{u}^{-1/2}du$, whose tail ${\int }_1^{\infty }{u}^{-1/2}du$ has $p=\frac{1}{2}<1$ and therefore diverges. A factor growing to infinity is not enough — it must grow like a power $>1$ of $\ln x$ (as $(\ln x{)}^2$ did).