3.9 Answer

3.9

1.

Note that we use the following symbols.

$$f(a +0) = \lim_{x \to a+}f(x), \ f(a-0) = \lim_{x \to a-}f(x)$$

(a) To integrate $\int_{0}^{1}\frac{dx}{\sqrt{1-x}}$, we need to know $f(x) = \frac{1}{\sqrt{1-x}}$ is continuous on $[0,1)$.

$$\int_{0}^{1}\frac{dx}{\sqrt{1-x}} = \int_{0}^{1-}\frac{dx}{\sqrt{1-x}}$$

Then we let $t = \sqrt{1-x}$. Then $t^2 = 1-x$ and $2tdt = -dx$．Note that

$$\begin{array}{l\vert lll} x&0 &\rightarrow & 1-0\\ \hline t&1 &\rightarrow & 0+ \end{array}$$

Then

$$\int_{0}^{1-}\frac{dx}{\sqrt{1-x}} = \int_{1}^{0+}{\frac{-2tdt}{t}}$$

$$= \int_{1}^{0+}(-2)\ dt \ \ (t )$$

$$= -2t\mid_{1}^{0+} = -2(\lim_{t \to 0+}t - 1)= 2$$

(b) $f(x) = \frac{1}{x}$ is continuous on $(0,1]$. Then

$$\int_{0}^{1}\frac{1}{x}\ dx = \int_{0+}^{1}\frac{1}{x}\ dx$$

Thus,

$$\int_{0+}^{1}\frac{1}{x}\ dx = \log\vert x\vert\mid_{0+}^{1} = \log{1} - \lim_{x \to 0+}\log\vert x\vert = \infty$$

(c) $f(x) = \log{x}$ is continuous on $(0,1]$. Then

$$\int_{0}^{1}\log{x}\ dx = \int_{0+}^{1}\log{x}\ dx$$

Using the integration by parts, we have

$$\left(\begin{array}{ll} f(x) = \log{x} & g'(x) = 1\\ f'(x) = \frac{1}{x} & g(x) = x \end{array}\right)$$

Thus,

$$\int_{0+}^{1}\log{x}\ dx = x\log{x}\mid_{0+}^{1} - \int_{0+}^{1}dx$$

$$= (x\log{x} - x)\mid_{0+}^{1}$$

$$= \log{1} - 1 - \lim_{x \to 0+}(x\log{x} - x)$$

$$= -1 - \lim_{x \to 0+}\frac{\log{x}}{\frac{1}{x}}$$

$$= -1 - \lim_{x \to 0+}\frac{1/x}{-1/x^2} \ \ ( )$$

$$= -1 - \lim_{x \to 0+}(-x) = -1$$

2.

(a) ， $f(x) = xe^{-x}$ is continuous on $[0,\infty)$. Then

$$\int_{0}^{\infty}xe^{-x}\ dx = \int_{0}^{\infty-}xe^{-x}\ dx$$

Using the integration by parts, we have

$$\int{udv} = uv - \int{vdu}$$

$$\left(\begin{array}{ll} u = x & dv = e^{-x}\\ du = dx & v = -e^{-x} \end{array}\right)$$

Thus,

$$\int_{0}^{\infty-}xe^{-x}\ dx = -xe^{-x}\mid_{0}^{\infty-} + \int_{0}^{\infty-}e^{-x}\ dx$$

$$= \lim_{x \to \infty-}(-xe^{-x} - e^{-x})\mid_{0}^{\infty-}$$

$$= \lim_{x \to \infty-}(\frac{-x}{e^{x}})$$

$$= \lim_{x \to \infty}\frac{-1}{e^{x}} = 0 \ ( )$$

(b) $f(x) = \frac{1}{x(\log{x})^{\alpha}}$ is continuous on $[2, \infty)$.

$$\int_{2}^{\infty}\frac{1}{x(\log{x})^{\alpha}}\ dx = \int_{2}^{\infty-}\frac{1}{x(\log{x})^{\alpha}}\ dx$$

Let $t = \log{x}$. Then $dt = \frac{1}{x}dx$．Note that

$$\begin{array}{l\vert lll} x&2&\rightarrow&\infty-\\ \hline t&log{2}&\rightarrow&\infty- \end{array}$$

Then

$$\int_{2}^{\infty-}\frac{1}{x(\log{x})^{\alpha}}\ dx = \int_{\log{2}}^{\infty-}\frac{1}{t^{-\alpha}}dt$$

$$= \left\{\begin{array}{ll} \frac{t^{-\alpha + 1}}{1 - \alpha} \mid_... ...\\ \log{\vert t\vert}\mid_{\log{2}}^{\infty-}, & \alpha = 1 \end{array}\right.$$

Note that

$$\lim_{x \to \infty-}{\frac{t^{-\alpha + 1}}{1 - \alpha}} = \left\{\begin{array}{ll} 0, & \alpha > 1\\ \infty, & \alpha < 1 \end{array}\right.$$

Then we have

$$\int_{2}^{\infty-}\frac{1}{x(\log{x})^{\alpha}}\ dx = \left\{\begin{array}{ll} 0, & \alpha > 1\\ \infty, & \alpha \leq 1 \end{array}\right.$$

(c) $f(x) = \frac{1}{x^{\alpha}\log{x}}$ is continuous on $[2, \infty)$. Then

$$\int_{2}^{\infty}\frac{1}{x^{\alpha}\log{x}}\ dx = \int_{2}^{\infty-}\frac{1}{x^{\alpha}\log{x}}\ dx$$

Now

$$x^{\alpha}f(x) = \frac{1}{\log{x}} \leq \frac{1}{\log{2}}$$

implies that

$$\int_{2}^{\infty-}f(x)\ dx \leq \frac{1}{\log{2}}\int_{2}^{\infty-}\frac{1}{x^{\alpha}}\ dx$$

Note that $\int_{2}^{\infty-}\frac{1}{x^{\alpha}}\ dx$ converges at $\alpha > 1$. For $\alpha = 1$，we let $t = \log{x}$. Then $dt = \frac{1}{x}dx$．Note that

$$\begin{array}{l\vert lll} x&2&\rightarrow&\infty-\\ \hline t&log{2}&\rightarrow&\infty- \end{array}$$

$$\int_{2}^{\infty}\frac{1}{x\log{x}}\ dx = \int_{\log{2}}^{\infty-}\frac{dt}{t}$$

$$= \log\vert t\vert\mid_{\log{2}}^{\infty-} = \infty$$

For $\alpha < 1$ and $2 < x < \infty$, $\frac{1}{x^{\alpha}} > \frac{1}{x}$ implies that

$$\int_{2}^{\infty}\frac{1}{x^{\alpha}\log{x}}\ dx > \int_{2}^{\infty}\frac{1}{x\log{x}}\ dx$$

Therefore, $\int_{2}^{\infty}\frac{1}{x^{\alpha}\log{x}}\ dx$ diverges．

3.

(a)

$$\cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots +$$

implies that．

$$\cos{x} \leq 1 - \frac{x^{2}}{2}$$

Thus,

$$\frac{1}{\sqrt{\cos{x}}} < \frac{1}{\sqrt{1 - \frac{x^2}{2}}}$$

Now if we can show

$$\int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{1 - \frac{x^2}{2}}}\ dx < \infty$$

then the integral converges.

$$\int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{1 - \frac{x^2}{2}}}\ dx = \sqrt{2}\int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{2 - x^2}}\ dx$$

$$= \sqrt{2}\sin{(\frac{x}{2})}\mid_{0}^{\frac{\pi}{2}} = \sqrt{2}\sin^{-1}{(\frac{\pi}{2\sqrt{2}})} < \infty$$

(b) $f(x)$ is continuous on $(0,1]$．Let $t = \sqrt{x}$. Then $t^2 = x$ and $2t dt = dx$．Note that

$$\begin{array}{l\vert lll} x&0+&\rightarrow&1\\ \hline t&0+&\rightarrow&1 \end{array}$$

Then

$$\int_{0+}^{1}\frac{\log{x}}{\sqrt{x}}\ dx = \int_{0+}^{1}\frac{\log{t^2}}{t}(2tdt) = 4\int_{0+}^{1}\log{t}\ dt$$

$$= 4[t\log{t} - t\mid_{0+}^{1} = 4[\log{1} - 1 - \lim_{t \to 0+}(t\log{t} - t)]$$

$$= 4[-1 - \lim_{t \to 0+}(\frac{\log{t} -1}{\frac{1}{t}})]$$

$$4[-1 - \lim_{t \to 0+}\frac{1/t}{-1/t^2} ] = -4$$