9.2 Workshop Solutions 2

True or False? (i) False. For example, let $f(x)=\frac{1}{x}$, then $\int_{1}^{R}\frac{dx}{x}=\left[\log x\right]_{1}^{R}=\log R\rightarrow\infty$ as $R\rightarrow\infty$.

(ii) False, e.g. $f(x)=\frac{1}{x}$. (See 1.54 in the notes.)

(iii) True, by the comparison test, since $\left|\frac{e^{-x}}{x^{2}}\right|\leq\frac{1}{x^{2}}$ for $x\geq 1$, so by the comparison test (1.57) $\int_{1}^{\infty}\frac{e^{-x}}{x^{2}}\,dx$ converges and has absolute value less than or equal to $\int_{1}^{\infty}\frac{dx}{x^{2}}=1$.

(iv) False. All we can say is that $f$ is a function of $y$ only. For example, we could have $f(x,y)=y$.

W2.1. i) We combine the logs to get

$${{1}\over{2}}\log(R^{2}+2)-\log(3R+1)={{1}\over{2}}\log{{R^{2}+2}\over{(3R+1)^{2}}}={{1}\over{2}}\log{{R^{2}+2}\over{9R^{2}+6R+1}}\rightarrow{{1}\over{2}}\log{{1}\over{9}}$$

as $R\rightarrow\infty$. Thus the limit of the expression as $R\rightarrow\infty$ is $-\log 3$.

ii) In this case, if we combine logs then we obtain $\log(R^{3}-3)-2\log(R-1)=\log\frac{R^{3}-3}{(R-1)^{2}}=\log{R^{3}-3}{R^{2}-2R+1}$. We have to cancel the same power on top and bottom, so this is equal to $\log\frac{R-3/R^{2}}{1-2/R+1/R^{2}}$. The expression inside the $\log$ tends to $\infty$ as $R\rightarrow\infty$, and therefore this expression diverges.

iii) In this case we combine everything into a single logarithm to obtain $\log\frac{(4R^{3}-1)^{2}}{(7R^{5}-2)(3R-2)}$. The highest degree term in $R$ on both top and bottom is $R^{6}$, hence this logarithm equals

$$\log\frac{(4-1/R^{3})^{2}}{(7-2/R^{5})(3-2/R)}=\log\frac{4^{2}}{21}=\log\frac{16}{21}.$$

iv) We have $\frac{R+1}{R^{2}-1}=\frac{1}{R-1}$ and hence the expression simplifies to $\frac{1}{R-1}\log(R-1)$. Now we can replace $R$ by $R+1$ without affecting convergence or divergence of this expression, so the question simplifies further to finding the limit of $\frac{1}{R}\log R=-\frac{1}{R}\log\frac{1}{R}$ as $R\rightarrow\infty$. By 4.23.1 in MATH101, $x\log x\rightarrow 0$ as $x\rightarrow 0+$, so substituting $x=\frac{1}{R}$ we obtain that the limit of the function is $0$.

The technically correct way to do these is to simplify,

then take limits at the last step.

W2.2. (i) We use partial fractions, and find that

$${{1}\over{(x+3)(5x+1)}}={{A}\over{x+3}}+{{B}\over{5x+1}}$$

when

$$1=A(5x+1)+B(x+3),$$

which gives the simultaneous equations for the coefficients

$$\begin{matrix}x:0=5A+B&&1=-14A&&A=-1/14\cr 1:1=A+3B&&0=5A+B&&B=5/14.\cr\end{matrix}$$

The integral to consider is

$$\int_{0}^{R}{{dx}\over{(x+3)(5x+1)}}=\int_{0}^{R}\Bigl({{-1/14}\over{x+3}}+{{5/14}\over{5x+1}}\Bigr)dx=\Bigl[-{{1}\over{14}}\log(x+3)+{{1}\over{14}}\log(5x+1)\Bigr]_{0}^{R}$$

$$={{1}\over{14}}\Bigl(-\log(R+3)+\log(5R+1)\Bigr)-{{1}\over{14}}\Bigl(-\log 3+\log 1\Bigr)$$

$$={{1}\over{14}}\log{{5R+1}\over{R+3}}+{{1}\over{14}}\log 3\rightarrow{{1}\over{14}}\log 5+{{1}\over{14}}\log 3\;\;\mbox{as}\;\;R\rightarrow\infty$$

since $(5R+1)/(R+3)\rightarrow 5$.

Hence the integral converges and

$$\int_{0}^{\infty}{{dx}\over{(x+3)(5x+1)}}={{1}\over{14}}\log 15.$$

(ii) In this case there are factors involving $1$ and $x^{2}$, but not $x$ itself, so the relevant partial fraction decomposition is simply

$${{1}\over{x^{2}(4+x^{2})}}={{A}\over{x^{2}}}+{{B}\over{4+x^{2}}},$$

which reduces to

$$1=A(4+x^{2})+Bx^{2}$$

and hence the simultaneous equations for the coefficients

$$\begin{matrix}x^{2}:0=A+B&&A=1/4\cr 1:1=4A&&B=-1/4.\cr\end{matrix}$$

Hence we consider the integral

$$\int_{2}^{R}{{dx}\over{x^{2}(4+x^{2})}}=\int_{2}^{R}\Bigl({{1/4}\over{x^{2}}}-{{1/4}\over{4+x^{2}}}\Bigr)dx=\Bigl[-{{1}\over{4x}}-{{1}\over{8}}\tan^{-1}{{x}\over{2}}\Bigr]_{2}^{R}$$

$$=-{{1}\over{4R}}-{{1}\over{8}}\tan^{-1}R/2-\Bigl({{-1}\over{8}}-{{1}\over{8}}\tan^{-1}1\Bigr)=-{{1}\over{4R}}-{{1}\over{8}}\tan^{-1}R/2+{{1}\over{8}}+{{\pi}\over{32}}\rightarrow-{{\pi}\over{16}}+{{1}\over{8}}+{{\pi}\over{32}}$$

as $R\rightarrow\infty$. Hence the integral converges to

$$\int_{2}^{\infty}{{dx}\over{x^{2}(4+x^{2})}}={{1}\over{8}}-{{\pi}\over{32}}.$$

W2.3. By the answer to W1.5(iv), we have

$$\int_{2}^{R}\frac{26}{f(x)}dx=\left[\frac{1}{2}\log\left(\frac{x^{2}-2x+10}{x^{2}-4x+8}\right)-\frac{2}{3}\tan^{-1}\left(\frac{x-1}{3}\right)+\frac{3}{2}\tan^{-1}\left(\frac{x-2}{2}\right)\right]_{2}^{R}$$

$$=\frac{1}{2}\log\left(\frac{R^{2}-2R+10}{R^{2}-4R+8}\right)-\frac{2}{3}\tan^{-1}\left(\frac{R-1}{3}\right)+\frac{3}{2}\tan^{-1}\left(\frac{R-2}{2}\right)-\frac{1}{2}\log\frac{10}{4}+\frac{2}{3}\tan^{-1}\frac{1}{3}-\frac{3}{2}\tan^{-1}0.$$

Since $\frac{R^{2}-2R+10}{R^{2}-4R+8}=\frac{1-2/R+10/R^{2}}{1-4/R+8/R^{2}}\rightarrow\frac{1}{1}=1$ as $R\rightarrow\infty$, the first term tends to zero. Further, $\tan^{-1}\left(\frac{R-1}{3}\right)$ and $\tan^{-1}\left(\frac{R-2}{2}\right)$ both tend to $\frac{\pi}{2}$ as $R\rightarrow\infty$, so the integral converges to

$$(\frac{3}{2}-\frac{2}{3})\frac{\pi}{2}-\frac{1}{2}\log\frac{10}{4}+\frac{2}{3}\tan^{-1}\frac{1}{3}=\frac{5\pi}{12}+\log\frac{2}{\sqrt{10}}+\frac{2}{3}\tan^{-1}\frac{1}{3}.$$

W2.4. i) This is an improper integral since $\frac{1}{\sqrt{x}}\rightarrow\infty$ as $x\rightarrow 0+$. Thus we consider $\int_{\delta}^{1}\frac{dx}{\sqrt{x}}=\left[2\sqrt{x}\right]_{\delta}^{1}=2(1-\sqrt{\delta})\rightarrow 2$ as $\delta\rightarrow 0+$. Therefore the integral converges to $2$. (This is just a special case of Example 1.54 in the notes.)

ii) This is an improper integral as $\sqrt{4-x^{2}}\rightarrow 0$ as $x\rightarrow 2-$. To evaluate this integral we make the substitution $x=2\sin t$. Then the limits of integration change as: $t=0$ when $x=0$, $t=\sin^{-1}(1-\frac{\delta}{2})$ when $x=2-\delta$. Further, $\frac{dx}{dt}=2\cos t$ and $\cos t$ is positive for this range of values of $t$, so $\sqrt{4-x^{2}}=\sqrt{4-4\sin^{2}t}=2\cos t$. Thus

$$\int_{0}^{2-\delta}\frac{dx}{\sqrt{4-x^{2}}}=\int_{0}^{\sin^{-1}(1-\frac{\delta}{2})}\frac{2\cos t\,dt}{2\cos t}=\sin^{-1}(1-\frac{\delta}{2})\rightarrow\sin^{-1}1=\frac{\pi}{2}$$

as $\delta\rightarrow 0+$.

Taking the bounds $0\leq x\leq 2-\delta$ rather than $0\leq x\leq 2$ is the formally correct way to check this integral.

iii) This is an improper integral as $\sqrt{x^{2}-4}\rightarrow 0$ as $x\rightarrow 2+$. To evaluate the integral we make the substitution $x=2\cosh t$. Then the bounds of integration change as: $t=\cosh^{-1}(1+\frac{\delta}{2})$ when $x=2+\delta$; $t=\cosh^{-1}2$ when $x=4$. Also, $\frac{dx}{dt}=2\sinh t$ and $\sinh t$ is positive for this range of values of $t$, so that $\sqrt{x^{2}-4}=\sqrt{4\sinh^{2}t}=2\sinh t$. Therefore

$$\int_{2+\delta}^{4}\frac{dx}{\sqrt{x^{2}-4}}=\int_{\cosh^{-1}(1+\frac{\delta}{2})}^{\cosh^{-1}2}\frac{2\sinh t\,dt}{2\sinh t}=(\cosh^{-1}2-\cosh^{-1}(1+\frac{\delta}{2}))\rightarrow\cosh^{-1}2$$

as $\delta\rightarrow 0+$.

iv) This is an improper integral as $1+\cos x=0$ when $x=\pi$. Thus $\frac{1}{1+\cos x}\rightarrow\infty$ as $x\rightarrow\pi$.

This is a rational function of $\cos x$ so the standard way to evaluate the integral is via the substitution $t=\tan\frac{x}{2}$. However, a quicker way to determine this integral is by the observation: $1+\cos x=1+(2\cos^{2}\frac{x}{2}-1)=2\cos^{2}\frac{x}{2}$. Therefore

$$\int_{0}^{\pi-\delta}\frac{dx}{1+\cos x}=\int_{0}^{\pi-\delta}\frac{1}{2}\sec^{2}\frac{x}{2}\,dx=\left[\tan\frac{x}{2}\right]_{0}^{\pi-\delta}=\tan\frac{\pi-\delta}{2}\rightarrow\infty$$

as $\delta\rightarrow 0+$. Thus the integral diverges.

v) This is an improper integral for the same reason as in (iv). Furthermore, in order for the integral $\int_{0}^{2\pi}\frac{dx}{1+\cos x}$ to converge, we need both $\int_{0}^{\pi}\frac{dx}{1+\cos x}$ and $\int_{\pi}^{2\pi}\frac{dx}{1+\cos x}$ to converge. Therefore the integral diverges.

This example shows that one should be careful about points of discontinuity inside the range of integration. A naive approach to this question would produce the answer $\int_{0}^{2\pi}\frac{dx}{1+\cos x}=\left[\tan\frac{x}{2}\right]_{0}^{2\pi}=0$, which is certainly not right!

W2.5. The Laplace transform is

$$F(s)=\int_{0}^{\infty}e^{-sx}f(x)dx.$$

When $f(x)=\cosh ax$ and $s>a>0$, we can calculate

$$\int_{0}^{R}e^{-sx}\cosh ax\,dx=\int_{0}^{R}e^{-sx}{{1}\over{2}}\bigl(e^{ax}+e^{-ax}\bigr)dx={{1}\over{2}}\int_{0}^{R}\bigl(e^{-(s-a)x}+e^{-(s+a)x}\bigr)dx$$

$$={{1}\over{2}}\Bigl[-{{e^{-(s-a)x}}\over{s-a}}-{{e^{-(s+a)x}}\over{s+a}}\Bigr]_{0}^{R}={{1}\over{2}}\Bigl[{{1}\over{s-a}}+{{1}\over{s+a}}-{{e^{-(s-a)R}}\over{s-a}}-{{e^{-(s+a)R}}\over{s+a}}\Bigr]\rightarrow{{1}\over{2}}\Bigl[{{1}\over{s-a}}+{{1}\over{s+a}}\Bigr]$$

as $R\rightarrow\infty$. Hence

$$F(s)={{s}\over{s^{2}-a^{2}}}\qquad(s>a).$$

Whereas the calculation by integration by parts as in the trigonometric integrals is possible, it is much more painful. It is also possible to use the fact that the Laplace transform of $\sinh ax$ is $\frac{a}{s^{2}-a^{2}}$ and Prop. 6.42 in the notes.

W2.6. We have $a^{x}=e^{x\log a}$. Thus

$$\int_{0}^{R}a^{x}\,dx=\int_{0}^{R}e^{x\log a}\,dx=\left[\frac{1}{\log a}e^{x\log a}\right]_{0}^{R}=\frac{1}{\log a}\left(a^{R}-a^{0}\right).$$

Now $a^{R}\rightarrow 0$ as $R\rightarrow\infty$ and $a^{0}=1$, so $\int_{0}^{\infty}a^{x}\,dx=-\frac{1}{\log a}$.

W2.7. (i) Let $u=x^{2}/2$, so that ${{du}\over{dx}}=x$, and

$$\begin{matrix}x&|&a&R\cr u&|&a^{2}/2&R^{2}/2\cr\end{matrix}$$

and substitute this into the integral to obtain

$$\int_{0}^{R}xe^{-x^{2}/2}dx=\int_{a^{2}/2}^{R^{2}/2}e^{-u}du=\Bigl[-e^{-u}\Bigr]_{a^{2}/2}^{R^{2}/2}=e^{-a^{2}/2}-e^{-R^{2}/2}\rightarrow e^{-a^{2}/2}$$

as $R\rightarrow\infty$; hence

$$\int_{a}^{\infty}xe^{-x^{2}/2}dx=e^{-a^{2}/2}.$$

(ii) We write $e^{-x^{2}/2}=x^{-1}\bigl(xe^{-x^{2}/2}\bigr)$ and integrate by parts to obtain

$$\int_{a}^{R}e^{-x^{2}/2}dx=\int_{a}^{R}x^{-1}\bigl(xe^{-x^{2}/2}\bigr)dx=\bigl[-x^{-1}e^{-x^{2}/2}\bigr]_{a}^{R}-\int_{a}^{R}x^{-2}e^{-x^{2}/2}dx.$$

We have $R^{-1}e^{-R^{2}/2}\rightarrow 0$ as $R\rightarrow\infty$, and the integral $\int_{a}^{\infty}x^{-2}e^{-x^{2}/2}dx$ converges since $\int_{a}^{\infty}x^{-2}dx$ converges; hence we can let $R\rightarrow\infty$ in the preceding identity and obtain

$$\int_{a}^{\infty}e^{-x^{2}/2}dx={{1}\over{a}}e^{-a^{2}/2}-\int_{a}^{\infty}{{1}\over{x^{2}}}e^{-x^{2}/2}dx.$$

Hermite showed that the integral $\int_{a}^{\infty}e^{-x^{2}/2}dx$ cannot be expressed in closed form in terms of elementary functions, where by ‘closed form’ we mean that no limits or other integrals are involved. As this integral arises in many applications, such as to the normal random variable in statistics, it is important to have techniques for estimating the numerical value of the integral.

W2.8. (i) If $f(x,y)=\sin x\cosh y$ then $\frac{\partial f}{\partial x}=\cos x\cosh y$ and $\frac{\partial f}{\partial y}=\sin x\sinh y$.

(ii) If $g(x,y)=\cos x\sinh y$ then ${{\partial g}\over{\partial x}}=-\sin x\sinh y$ and ${{\partial g}\over{\partial y}}=\cos x\cosh y$.

W2.9. (i) The various first-order partial derivative of $f=z\sinh(yz^{3}+x^{2})$ are

$${{\partial f}\over{\partial x}}=2xz\cosh(yz^{3}+x^{2}),\;{{\partial f}\over{\partial y}}=z^{4}\cosh(yz^{3}+x^{2}),\;{{\partial f}\over{\partial z}}=\sinh(yz^{3}+x^{2})+3yz^{3}\cosh(yz^{3}+x^{2}).$$

(ii) If

$$g(x,y,z)=e^{x+2y+3z}$$

then

$${{\partial g}\over{\partial x}}=e^{x+2y+3z},\;{{\partial g}\over{\partial y}}=2e^{x+2y+3z},{{\partial g}\over{\partial z}}=3e^{x+2y+3z}.$$

W2.10. The function $u(x,t)=\sin(x^{2}-t)$ has

$${{\partial u}\over{\partial x}}=2x\cos(x^{2}-t)\quad{\hbox{and}}\quad{{\partial u}\over{\partial t}}=-\cos(x^{2}-t),$$

hence

$${{\partial u}\over{\partial x}}+2x{{\partial u}\over{\partial t}}=0.$$

W2.11. Set $f=x+y^{5}+z^{7}$, so that $w=f^{6}$. Note that $\frac{\partial f}{\partial x}=1$, $\frac{\partial f}{\partial y}=5y^{4}$ and $\frac{\partial f}{\partial z}=7z^{6}$. By the product rule,

$$\frac{\partial}{\partial x}\left(f^{6}\right)=6f^{5}\frac{\partial f}{\partial x}=6f^{5}$$

Now $w_{xy}=(w_{x})_{y}=\frac{\partial}{\partial y}\left(w_{x}\right)=6\frac{\partial}{\partial y}\left(f^{5}\right)$. Using the product rule again, $\frac{\partial}{\partial y}\left(f^{5}\right)=5f^{4}\frac{\partial f}{\partial y}=25y^{4}f^{4}$. Hence $w_{xy}=150y^{4}f^{4}$.

Finally, $w_{xyz}=(w_{xy})_{z}$, so $w_{xyz}=\frac{\partial}{\partial z}\left(150y^{4}f^{4}\right)=150y^{4}\frac{\partial}{\partial z}\left(f^{4}\right)=150y^{4}.4f^{3}\frac{\partial f}{\partial z}=600y^{4}f^{3}.7z^{6}$. Thus $w_{xyz}=4200y^{4}z^{6}(x+y^{5}+z^{7})^{3}$.

W2.12. (i) Easy way (via implicit differentiation) Differentiating by $x$ and using the product rule, we obtain:

$$3x^{2}y+x^{3}\frac{dy}{dx}+2y\frac{dy}{dx}=0,$$

hence, by rearranging, $(x^{3}+2y)\frac{dy}{dx}=-3x^{2}y$, so $\frac{dy}{dx}=-\frac{3x^{2}y}{x^{3}+2y}$.

Second way (similar to 2.9-10 in the notes) We can rearrange the equation to obtain: $(y+\frac{x^{3}}{2})^{2}=2+\frac{x^{6}}{4}$, hence $y=\frac{-x^{3}\pm\sqrt{8+x^{6}}}{2}$. Now we can differentiate to obtain $\frac{dy}{dx}=\frac{1}{2}\left(-3x^{2}\pm 3x^{5}(8+x^{6})^{\frac{-1}{2}}\right)$.

(ii) We take the partial derivatives with respect to $S$ on the left- and right-hand sides. Then we obtain:

$$re^{rS}=2a\frac{\partial a}{\partial S}S^{2}+2a^{2}S$$

so, rearranging, we obtain: $2aS^{2}\frac{\partial a}{\partial S}=re^{rS}-2a^{2}S$, whence $\frac{\partial a}{\partial S}=\frac{re^{rS}-2a^{2}S}{2aS^{2}}$.

(iii) Once again, we take partial derivatives. Looking at each term in turn: on the left-hand side, $\frac{\partial}{\partial q}(pm)=m\frac{\partial p}{\partial q}$, $\frac{\partial}{\partial q}(pq)=p+q\frac{\partial p}{\partial q}$ and $\frac{\partial}{\partial q}(qm)=m$. On the right-hand side we obtain $\frac{\partial}{\partial q}(\cos(pqm))=-\frac{\partial}{\partial q}(pqm)\sin(pqm)=-(pm+mq\frac{\partial p}{\partial q})\sin(pqm)$. Thus

$$m\frac{\partial p}{\partial q}+p+q\frac{\partial p}{\partial q}+m=-\left(pm+mq\frac{\partial p}{\partial q}\right)\sin(pqm)$$

and so, rearranging, we get $(m+q+mq\sin(pqm))\frac{\partial p}{\partial q}=-(m+p+mp\sin(pqm))$. Rearranging, we obtain:

$$\frac{\partial p}{\partial q}=-\frac{m+p+mp\sin(pqm)}{m+q+mq\sin(pqm)}.$$

W2.13. (i) With $f(x,y)=y/x$ we have first-order partial derivatives

$$f_{x}=-y/x^{2},\quad{\hbox{and}}\quad f_{y}=1/x,$$

and second-order partial derivatives

$$f_{xx}=2y/x^{3},\quad f_{yy}=0,\quad f_{xy}=-1/x^{2}.$$

(ii) Recall that ${{d}\over{du}}\tan^{-1}u=1/(1+u^{2})$. So with $g(x,y)=\tan^{-1}(y/x),$ we have, by the chain rule and the results of (i), the first-order partial derivatives

$$g_{x}={{1}\over{1+y^{2}/x^{2}}}(-y/x^{2})={{-y}\over{x^{2}+y^{2}}},\;\;g_{y}={{1}\over{1+y^{2}/x^{2}}}(1/x)={{x}\over{x^{2}+y^{2}}},$$

and second-order partial derivatives

$$g_{xx}={{2xy}\over{(x^{2}+y^{2})^{2}}},\quad g_{yy}={{-2xy}\over{(x^{2}+y^{2})^{2}}},$$

$$g_{xy}={{\partial}\over{\partial y}}\Bigl(-y(x^{2}+y^{2})^{-1}\Bigr)=-(x^{2}+y^{2})^{-1}+2y^{2}(x^{2}+y^{2})^{-2}={{y^{2}-x^{2}}\over{(x^{2}+y^{2})^{2}}}.$$

In the final step it is slightly easier to use the product rule than the quotient rule.

W2.14. (i) By the quotent rule, we have

$${{d}\over{ds}}{\hbox{sech}}\,s={{d}\over{ds}}{{1}\over{\cosh s}}=-{{\sinh s}\over{\cosh^{2}s}}=-{\hbox{sech}}\,s\tanh s.$$

(ii) With $f(s)=-2^{-1}{\hbox{sech}}^{2}(s/2)$, we have

$$f^{\prime}(s)=(-2^{-1})2{\hbox{sech}}\,(s/2)(-1/2){\hbox{sech}}(s/2)\tanh(s/2)=2^{-1}{\hbox{sech}}^{2}(s/2)\tanh(s/2),$$

so

$$f^{\prime}(s)^{2}=4^{-1}{\hbox{sech}}\,^{4}(s/2)\tanh^{2}(s/2),$$

while

$$f^{2}(2f+1)=4^{-1}{\hbox{sech}}^{4}(s/2)(1-{\hbox{sech}}^{2}(s/2))=4^{-1}{\hbox{sech}}^{4}(s/2)\tanh^{2}(s/2);$$

hence

$$(f^{\prime})^{2}=f^{2}(2f+1).$$