Evaluate the Integral or State That It Diverges: Six Improper Integrals

Answer

Every part is an improper integral, so each one starts the same way: replace the infinite limit or the singular point with a limit variable, integrate over the interval where the integrand is continuous, and then decide whether that limit exists. A finite limit is the value of the integral; an infinite or non-existent limit means the integral diverges.

(a) $\int_{1}^{\infty} x^{-e}\,dx$ converges to $\frac{1}{e-1}$

$\int_{1}^{\infty}x^{-e}\,dx=\lim_{t\to\infty}\int_{1}^{t}x^{-e}\,dx=\lim_{t\to\infty}\left[\frac{x^{1-e}}{1-e}\right]_{1}^{t}=\lim_{t\to\infty}\frac{t^{1-e}-1}{1-e}.$

Since $e\approx 2.718>1$, the exponent $1-e$ is negative, so $t^{1-e}$ tends to $0$. The limit is $\frac{0-1}{1-e}=\frac{1}{e-1}\approx 0.582$, and the integral converges.

This is the $p$-integral $\int_{1}^{\infty}x^{-p}dx$ with $p=e$. Over an infinite interval it converges exactly when $p>1$.

(b) $\int_{0}^{1} x^{-\pi}\,dx$ diverges

The integrand blows up at the lower endpoint, so the integral is defined by

$\int_{0}^{1}x^{-\pi}\,dx=\lim_{t\to 0^{+}}\left[\frac{x^{1-\pi}}{1-\pi}\right]_{t}^{1}=\lim_{t\to 0^{+}}\frac{1-t^{1-\pi}}{1-\pi}.$

Because $\pi\approx 3.142>1$, the exponent $1-\pi$ is negative and $t^{1-\pi}=t^{-(\pi-1)}$ grows without bound as $t\to 0^{+}$. The numerator tends to $-\infty$ while the denominator $1-\pi$ is negative, so the whole expression tends to $+\infty$. The integral diverges.

The direction of the inequality matters at a finite endpoint: $\int_{0}^{1}x^{-p}dx$ converges only when $p<1$, which is the opposite of the infinite-interval rule used in part (a).

(c) $\int_{0}^{1}\frac{dx}{\sqrt{1-x}}$ converges to $2$

$\int_{0}^{1}\frac{dx}{\sqrt{1-x}}=\lim_{t\to 1^{-}}\left[-2\sqrt{1-x}\right]_{0}^{t}=\lim_{t\to 1^{-}}\left(2-2\sqrt{1-t}\right)=2.$

The blow-up is of square-root type, which is integrable, so the integral converges to $2$.

(d) $\int_{0}^{1}\frac{dx}{1-x}$ diverges

$\int_{0}^{1}\frac{dx}{1-x}=\lim_{t\to 1^{-}}\left[-\ln(1-x)\right]_{0}^{t}=\lim_{t\to 1^{-}}\left(-\ln(1-t)\right)=+\infty.$

The integral diverges.

Parts (c) and (d) share the same singular endpoint, yet only (c) converges. The difference is the strength of the blow-up: $(1-x)^{-1/2}$ grows slowly enough to be integrable there, while $(1-x)^{-1}$ does not.

(e) $\int_{0}^{\infty} x e^{-x}\,dx$ converges to $1$

$\int_{0}^{\infty}xe^{-x}dx=\lim_{t\to\infty}\left[-(x+1)e^{-x}\right]_{0}^{t}=\lim_{t\to\infty}\left(1-(t+1)e^{-t}\right)=1,$

because the exponential decays faster than the linear factor grows. The integral converges to $1$.

(f) $\int_{0}^{1}\frac{\ln x}{x}\,dx$ diverges

Substituting $u=\ln x$, so that $du=\frac{dx}{x}$, turns the integrand into $u\,du$ with antiderivative $\frac{u^{2}}{2}=\frac{(\ln x)^{2}}{2}$. Therefore

$\int_{0}^{1}\frac{\ln x}{x}\,dx=\lim_{t\to 0^{+}}\left[\frac{(\ln x)^{2}}{2}\right]_{t}^{1}=\lim_{t\to 0^{+}}\left(0-\frac{(\ln t)^{2}}{2}\right)=-\infty.$

The integral diverges.

Summary

"Diverges" records that no finite limiting value exists. It does not mean the area is zero, and it does not mean the integral can be assigned some other finite value by symmetry.