Solutions to Problem Set 9

Math 211-03

9-21-2017

[Improper integrals]

1. Determine whether the improper integral $\displaystyle \int_0^\infty e^{-5
   x}\,dx$ converges or diverges.

If the integral converges, compute its exact value.

$$\int_0^\infty e^{-5 x}\,dx = \lim_{b \to \infty} \int_0^b e^{-5 x}\,dx = \lim_{b \to \infty} \left[-\dfrac{1}{5} e^{-5 x}\right]_0^b =$$

$$\lim_{b \to \infty} \left(-\dfrac{1}{5} e^{-5 b} + \dfrac{1}{5}\right) = 0 + \dfrac{1}{5} = \dfrac{1}{5}.\quad\halmos$$


2. Determine whether the improper integral $\displaystyle \int_0^\infty
   \dfrac{1}{x^2 + 4}\,dx$ converges or diverges.

If the integral converges, compute its exact value.

$$\int_0^\infty \dfrac{1}{x^2 + 4}\,dx = \lim_{b \to \infty} \int_0^b \dfrac{1}{x^2 + 4}\,dx = \lim_{b \to \infty} \left[\dfrac{1}{2} \tan^{-1} \dfrac{x}{2}\right]_0^b =$$

$$\lim_{b \to \infty} \left(\dfrac{1}{2} \tan^{-1} \dfrac{b}{2} - 0\right) = \dfrac{1}{2} \cdot \dfrac{\pi}{2} = \dfrac{\pi}{4}.\quad\halmos$$


3. Determine whether the improper integral $\displaystyle \int_1^\infty
   \dfrac{1}{\sqrt{3 x + 1}}\,dx$ converges or diverges.

If the integral converges, compute its exact value.

$$\int_1^\infty \dfrac{1}{\sqrt{3 x + 1}}\,dx = \lim_{b \to \infty} \int_1^b \dfrac{1}{\sqrt{3 x + 1}}\,dx = \lim_{b \to \infty} \left[\dfrac{2}{3} (3 x + 1)^{1/2}\right]_1^b = \lim_{b \to \infty} \left(\dfrac{2}{3} (3 b + 1)^{1/2} - \dfrac{4}{3}\right).$$

Since $\displaystyle \lim_{b \to
   \infty} \dfrac{2}{3} (3 b + 1)^{1/2} = \infty$ , the integral diverges. I may write

$$\int_1^\infty \dfrac{1}{\sqrt{3 x + 1}}\,dx = \infty.\quad\halmos$$


4. Determine whether the improper integral $\displaystyle \int_0^\infty \cos 5
   x\,dx$ converges or diverges.

If the integral converges, compute its exact value.

$$\int_0^\infty \cos 5 x\,dx = \lim_{b \to \infty} \int_0^b \cos 5 x\,dx = \lim_{b \to \infty} \left[\dfrac{1}{5} \sin 5 x\right]_0^b = \lim_{b \to \infty} \dfrac{1}{5} \sin 5 b.$$

$\displaystyle \lim_{b \to
   \infty} \dfrac{1}{5} \sin 5 b$ is undefined. Hence, the integral diverges.

Note: The integral is not "$\infty$ " or "$-\infty$ ", since it is not getting arbitrarily large and positive or arbitrarily large and negative.


5. Determine whether the improper integral $\displaystyle \int_{-\infty}^\infty
   e^{-x}\,dx$ converges or diverges.

If the integral converges, compute its exact value.

First, break up the original integral into two improper integrals, each having one infinite limit:

$$\int_{-\infty}^\infty e^{-x}\,dx = \int_{-\infty}^0 e^{-x}\,dx + \int_0^\infty e^{-x}\,dx.$$

(I divided the original integral at 0, but you could use any number.)

I'll do the integrals separately. If either one diverges, then the original integral diverges. Otherwise, if both converge, the value of the original integral is their sum.

$$\int_{-\infty}^0 e^{-x}\,dx = \lim_{b \to -\infty} \int_b^0 e^{-x}\,dx = \lim_{b \to -\infty} \left[-e^{-x}\right]_b^0 = \lim_{b \to -\infty} \left(-1 + e^{-b}\right).$$

Now

$$\lim_{b \to -\infty} e^{-b} = \infty.$$

Hence, $\displaystyle
   \int_{-\infty}^0 e^{-x}\,dx$ diverges.

I don't need to compute the second integral $\displaystyle \int_0^\infty
   e^{-x}\,dx$ . The original integral diverges.


6. Determine whether the improper integral $\displaystyle \int_0^1
   \dfrac{1}{x^{3/5}}\,dx$ converges or diverges.

If the integral converges, compute its exact value.

The function $f(x) =
   \dfrac{1}{x^{3/5}}$ is undefined at $x = 0$ .

$$\int_0^1 \dfrac{1}{x^{3/5}}\,dx = \lim_{b \to 0^+} \int_b^1 \dfrac{1}{x^{3/5}}\,dx = \lim_{b \to 0^+} \left[\dfrac{5}{2} x^{2/5}\right]_b^1 = \lim_{b \to 0^+} \left(\dfrac{5}{2} - \dfrac{5}{2} b^{2/5}\right) = \dfrac{5}{2} - 0 = \dfrac{5}{2}.\quad\halmos$$


7. Determine whether the improper integral $\displaystyle \int_0^1
   \dfrac{1}{x^{7/6}}\,dx$ converges or diverges.

If the integral converges, compute its exact value.

$$\int_0^1 \dfrac{1}{x^{7/6}}\,dx = \lim_{b \to 0^+} \int_b^1 \dfrac{1}{x^{7/6}}\,dx = \lim_{b \to 0^+} \left[\dfrac{-6}{x^{1/6}}\right]_b^1 = \lim_{b \to 0^+} \left(-6 + \dfrac{6}{b^{1/6}}\right).$$

Now $\displaystyle \lim_{b \to
   0^+} \dfrac{6}{b^{1/6}} = \infty$ , so

$$\int_0^1 \dfrac{1}{x^{7/6}}\,dx = \infty.$$

The integral diverges.


8. Determine whether the improper integral $\displaystyle \int_0^2
   \dfrac{1}{\sqrt{2 - x}}\,dx$ converges or diverges.

If the integral converges, compute its exact value.

$$\int_0^2 \dfrac{1}{\sqrt{2 - x}}\,dx \lim_{b \to 2^-} \int_0^b \dfrac{1}{\sqrt{2 - x}}\,dx = \lim_{b \to 2^-} \left[-2 \sqrt{2 - x}\right]_0^b = -2 \lim_{b \to 2^-} \left(\sqrt{2 - b} - \sqrt{2}\right) = 2 \sqrt{2}.\quad\halmos$$


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