Improper integrals and convergence tests

Type 1 — infinite limit. When the interval itself is unbounded, replace $\infty$ by a dummy variable $b$ and let $b\to\infty$ afterward:

where: $a$ is the fixed finite endpoint, $b$ is a temporary finite upper limit, and $f$ is continuous on $[a,\infty)$. The integral $\int_{-\infty}^b f\,dx$ is defined symmetrically, and $\int_{-\infty}^\infty f\,dx$ is split at any convenient point into two such pieces — both must converge independently for the whole thing to converge.

Type 2 — unbounded integrand. When $f$ itself misbehaves at an endpoint, say $f(x)\to\pm\infty$ as $x\to b^-$, the same trick applies to the integrand's domain instead of the interval:

where: $c$ is a point just short of the singularity $b$, and the limit is taken from inside $[a,b)$. A singularity at the lower limit or at an interior point is handled the same way, splitting the integral there so each piece has its bad behavior isolated at one endpoint.

The $p$-test. Two benchmark families decide convergence at a glance:

where: $p$ is a fixed real exponent, the first integral probes behavior as $x\to\infty$, and the second probes behavior as $x\to 0^+$. Notice the conditions are opposite in direction — that asymmetry is exactly what the figure below makes visible.

Comparison test. If $0\le f(x)\le g(x)$ for $x\ge a$, then a convergent $\int_a^\infty g$ forces $\int_a^\infty f$ to converge too (squeezed underneath), and a divergent $\int_a^\infty f$ forces $\int_a^\infty g$ to diverge (it can only be bigger). The limit comparison test is often more convenient: if $f,g>0$ and $\lim_{x\to\infty} f(x)/g(x)=L$ with $0<L<\infty$, then $\int_a^\infty f$ and $\int_a^\infty g$ share the same fate. Both forms carry over unchanged to Type 2 integrals, using $\int_0^1 x^{-p}dx$ as the benchmark instead.

Problem. Test $\displaystyle\int_1^\infty \frac{dx}{\sqrt{x^4+1}}$ for convergence using the comparison test, without attempting to evaluate it exactly.

Step 1. Look for the dominant behavior as $x\to\infty$. Inside the square root, $x^4$ dominates the $+1$ once $x$ is large, so $\sqrt{x^4+1}\approx x^2$ and the integrand behaves like $1/x^2$ — the natural benchmark is the $p$-integral with $p=2$.

Step 2. Make the comparison rigorous, not just approximate. Since $1\ge 0$, we have $x^4+1\ge x^4$ for every $x\ge 1$, so taking square roots (a valid, order-preserving step for positive quantities) gives $\sqrt{x^4+1}\ge x^2$. Inverting a positive inequality flips it, so

Step 3. Confirm the benchmark converges. By the derivation above with $p=2>1$, $\int_1^\infty x^{-2}dx$ converges, to the exact value $1/(2-1)=1$.

Step 4. Apply the comparison test. Our integrand is squeezed between $0$ and a convergent integrand's values, so by the comparison test $\displaystyle\int_1^\infty \frac{dx}{\sqrt{x^4+1}}$ converges — even though we never found (and do not need) its exact value.

Competitive stretch

Push the same idea to $\int_1^\infty \frac{dx}{\sqrt{x^6+x^2}}$: for large $x$, $x^6$ dominates, so $\sqrt{x^6+x^2}\ge x^3$ once $x\ge 1$ (since $x^6+x^2\ge x^6$), giving the bound $1/\sqrt{x^6+x^2}\le 1/x^3$. Since $p=3>1$ converges, the original integral converges too — the technique scales to any dominant power without extra machinery.

• Plugging $\infty$ directly into a formula instead of writing the limit $\lim_{b\to\infty}$ explicitly — always set the limit up first, evaluate second.
• Reversing the $p$-test direction: $\int_1^\infty x^{-p}dx$ needs $p>1$ (large exponent), while $\int_0^1 x^{-p}dx$ needs $p<1$ (small exponent) — these are opposite conditions and get swapped under pressure.
• Missing an interior singularity, such as $\int_{-1}^1 \frac{dx}{x^2}$ blowing up at $x=0$ inside the interval — you must split at the singularity and check both halves separately, not just the endpoints.
• Using the comparison test with the inequality pointing the wrong way — bounding your integrand above by a divergent integral, or below by a convergent one, proves nothing at all.

Ask the tutor: if you are unsure which benchmark power to compare against for a tricky integrand, ask the AI tutor to walk through picking the dominant term with you on a fresh example.