Improper Integrals

$\displaystyle\int_0^\infty xe^{-2x}\,dx = \lim_{R\to\infty}\int_0^R xe^{-2x}\,dx$.

Parts with $u = x$, $dv = e^{-2x}\,dx$: $F(x) = -\dfrac{x}{2}e^{-2x} - \dfrac{1}{4}e^{-2x}$.

$F(R) - F(0) = -\dfrac{R}{2}e^{-2R} - \dfrac{1}{4}e^{-2R} + \dfrac{1}{4}$.

$Re^{-2R} = \dfrac{R}{e^{2R}} \to 0$ (l'Hôpital) and $e^{-2R} \to 0$, so the limit is $\dfrac{1}{4}$.

$\displaystyle\lim_{R\to\infty}\int_3^R \frac{2}{x^2 - 1}\,dx$.

$\dfrac{2}{x^2 - 1} = \dfrac{1}{x - 1} - \dfrac{1}{x + 1}$, so $F(x) = \log(x - 1) - \log(x + 1) = \log\dfrac{x - 1}{x + 1}$.

$F(R) - F(3) = \log\dfrac{R - 1}{R + 1} - \log\dfrac{2}{4} = \log\dfrac{R - 1}{R + 1} + \log 2$.

$\dfrac{R - 1}{R + 1} \to 1$ and $\log 1 = 0$, so the limit is $\log 2$.

Split at $x = -2$: $\displaystyle\lim_{S\to-\infty}\int_S^{-2}\frac{dx}{x^2 + 4x + 5} + \lim_{R\to\infty}\int_{-2}^R\frac{dx}{x^2 + 4x + 5}$.

Complete the square: $x^2 + 4x + 5 = (x + 2)^2 + 1$, so $F(x) = \arctan(x + 2)$.

Left half: $\arctan 0 - \arctan(S + 2) = -\arctan(S + 2)$. Right half: $\arctan(R + 2) - \arctan 0 = \arctan(R + 2)$.

$-\arctan(S + 2) \to \dfrac{\pi}{2}$ and $\arctan(R + 2) \to \dfrac{\pi}{2}$. Both halves converge; they add to $\pi$.

$\displaystyle\lim_{R\to\infty}\int_1^R \frac{x}{x^2 + 3}\,dx$.

Substitute $u = x^2 + 3$: $F(x) = \tfrac{1}{2}\log(x^2 + 3)$.

$F(R) - F(1) = \tfrac{1}{2}\log(R^2 + 3) - \tfrac{1}{2}\log 4$.

$\log(R^2 + 3) \to \infty$, so there is no finite limit.

$\log x \to -\infty$ as $x \to 0^+$. The only bad point is the endpoint $x = 0$.

It is already an endpoint: nothing to split.

$\displaystyle\lim_{t\to 0^+}\int_t^1 \log x\,dx$.

Parts: $\int \log x\,dx = x\log x - x$. So $\int_t^1 \log x\,dx = -1 - t\log t + t \to -1$, since $t\log t \to 0$ (l'Hôpital).

$(x - 1)^{2/3} = 0$ at $x = 1$, which is inside $[0, 3]$.

$\displaystyle\int_0^1 \frac{dx}{(x - 1)^{2/3}} + \int_1^3 \frac{dx}{(x - 1)^{2/3}}$.

$\displaystyle\lim_{t\to 1^-}\int_0^t \frac{dx}{(x - 1)^{2/3}} + \lim_{s\to 1^+}\int_s^3 \frac{dx}{(x - 1)^{2/3}}$.

$F(x) = 3(x - 1)^{1/3}$. Left: $3(t - 1)^{1/3} - 3(-1) \to 3$. Right: $3\sqrt[3]{2} - 3(s - 1)^{1/3} \to 3\sqrt[3]{2}$. Both converge.

$x^2 - 4 = 0$ at $x = \pm 2$, and $x = 2$ is inside $[0, 3]$.

$\displaystyle\int_0^2 \frac{2x}{x^2 - 4}\,dx + \int_2^3 \frac{2x}{x^2 - 4}\,dx$.

Start with the left piece: $\displaystyle\lim_{t\to 2^-}\int_0^t \frac{2x}{x^2 - 4}\,dx$.

$F(x) = \log|x^2 - 4|$, so the left piece is $\log|t^2 - 4| - \log 4 \to -\infty$. One piece diverges, so the whole integral does.

Already $\dfrac{1}{x^p}$, with $p = 2k + 1$.

Improper only at $\infty$: the integrand is fine on $[1, \infty)$.

Converges exactly when $p > 1$.

$2k + 1 > 1$ means $k > 0$.

Already $\dfrac{1}{x^p}$, with $p = 3a$.

Improper at $0$ (when $3a > 0$).

Converges exactly when $p < 1$. For $p \le 0$ it is an ordinary integral, so it converges too.

$3a < 1$ means $a < \tfrac{1}{3}$.

$u = \log x$, $du = \dfrac{dx}{x}$; $x = e$ gives $u = 1$, and $u \to \infty$ as $x \to \infty$: $\displaystyle\int_1^\infty \frac{du}{u^p}$.

Improper only at $\infty$.

Converges exactly when $p > 1$.

Nothing left to solve: the parameter is $p$ itself.

Split at $1$: $\displaystyle\int_0^1 \frac{dx}{x^p} + \int_1^\infty \frac{dx}{x^p}$.

The first piece is improper at $0$ (when $p > 0$), the second at $\infty$.

The first needs $p < 1$; the second needs $p > 1$.

No $p$ is both below $1$ and above $1$.

The integrand is continuous on $[1, \infty)$; the only problem is $\infty$.

$\sin x \le 1$, so the top is at most $3$: try $g(x) = \dfrac{3}{x^2}$.

$\displaystyle\int_1^\infty \frac{3}{x^2}\,dx$ converges ($p = 2 > 1$).

$0 \le \dfrac{2 + \sin x}{x^2} \le \dfrac{3}{x^2}$ for $x \ge 1$.

Converges, by the comparison test.

Continuous on $[1, \infty)$; improper only at $\infty$.

For large $x$: $\dfrac{\sqrt{x} + 1}{x + 3} \approx \dfrac{\sqrt{x}}{x} = \dfrac{1}{\sqrt{x}}$. Take $g(x) = \dfrac{1}{x^{1/2}}$.

$\displaystyle\int_1^\infty \frac{dx}{x^{1/2}}$ diverges ($p = \tfrac{1}{2} \le 1$).

$\displaystyle\lim_{x\to\infty}\frac{f(x)}{g(x)} = \lim_{x\to\infty}\frac{x + \sqrt{x}}{x + 3} = 1$, which is finite and not $0$.

Diverges, by the limiting comparison test.

Infinite only at $x = 0$, which is already an endpoint.

Near $0$, $\cos x \approx 1$: try $g(x) = \dfrac{1}{x^{2/3}}$.

$\displaystyle\int_0^1 \frac{dx}{x^{2/3}}$ converges ($p = \tfrac{2}{3} < 1$).

On $(0, 1]$, $0 < \cos x \le 1$, so $0 \le \dfrac{\cos x}{x^{2/3}} \le \dfrac{1}{x^{2/3}}$.

Converges, by the comparison test.

Infinite at $x = 0$, and the domain reaches $\infty$. Split at $1$: $\displaystyle\int_0^1 + \int_1^\infty$.

Near $0$, $e^{-x} \approx 1$: $g_1 = \dfrac{1}{\sqrt{x}}$. For $x \ge 1$, $\sqrt{x} \ge 1$: $g_2 = e^{-x}$.

$\displaystyle\int_0^1 \frac{dx}{\sqrt{x}}$ converges ($p = \tfrac{1}{2} < 1$), and $\displaystyle\int_1^\infty e^{-x}\,dx = e^{-1}$ converges.

On $(0, 1]$: $0 \le \dfrac{e^{-x}}{\sqrt{x}} \le \dfrac{1}{\sqrt{x}}$. On $[1, \infty)$: $0 \le \dfrac{e^{-x}}{\sqrt{x}} \le e^{-x}$.

Both pieces converge by the comparison test, so the whole integral converges.