Proof of Dirichlet’s test for improper integrals

Let
[ F(X)=\int_a^X f(t),dt . ] By hypothesis, (F(X)) is bounded on ([a,\infty)). So there exists (M>0) such that
[ |F(X)|\le M \qquad \text{for all } X\ge a. ]

Since (\phi) is monotonic and (\phi(x)\to 0) as (x\to\infty), we may assume first that (\phi) is decreasing to (0). Then (\phi(x)\ge 0).
If (\phi) is increasing to (0), we apply the same argument to (-\phi), which is decreasing to (0).

For any (B>A\ge a), by Bonnet’s second mean value theorem for integrals, there exists (\xi\in[A,B]) such that
[ \int_A^B f(x)\phi(x),dx

\phi(A)\int_A^\xi f(x),dx . ] But
[ \int_A^\xi f(x),dx=F(\xi)-F(A). ] Hence
[ \left|\int_A^B f(x)\phi(x),dx\right|

\phi(A)|F(\xi)-F(A)| \le \phi(A)\bigl(|F(\xi)|+|F(A)|\bigr) \le 2M\phi(A). ]

Since (\phi(A)\to 0) as (A\to\infty), for every (\varepsilon>0) there exists (A_0\ge a) such that
[ A\ge A_0 \implies 2M\phi(A)<\varepsilon. ] Therefore, for all (B>A\ge A_0),
[ \left|\int_A^B f(x)\phi(x),dx\right|<\varepsilon. ] By Cauchy’s criterion for improper integrals,
[ \int_a^\infty f(x)\phi(x),dx ] is convergent at (\infty). (\square)

OR

Show that (\displaystyle \int_0^1 \log\sin x,dx) is convergent and evaluate it.

Near (x=0),
[ \log\sin x \sim \log x, ] and
[ \int_0^1 \log x,dx=[x\log x-x]_0^1=-1, ] so the integral converges at (0). At (x=1), (\log\sin x) is finite. Hence the integral is convergent.

For evaluation, use the standard Fourier expansion
[ \log\sin x=-\log 2-\sum_{n=1}^\infty \frac{\cos 2nx}{n},\qquad 0<x<\pi. ] Integrating from (0) to (\pi/2),
[ \int_0^{\pi/2}\log\sin x,dx

-\frac{\pi}{2}\log 2 -\sum_{n=1}^\infty \frac1n\int_0^{\pi/2}\cos 2nx,dx. ] But
[ \int_0^{\pi/2}\cos 2nx,dx

\left[\frac{\sin 2nx}{2n}\right]_0^{\pi/2}

\frac{\sin n\pi}{2n}=0. ] Therefore
[ \boxed{\int_0^{\pi/2}\log\sin x,dx=-\frac{\pi}{2}\log 2 }. ]

If the upper limit is literally (1) instead of (\pi/2), the value is
[ \int_0^1 \log\sin x,dx

-\log 2-\frac12\sum_{n=1}^\infty \frac{\sin 2n}{n^2}

-\log 2-\frac12\operatorname{Cl}_2(2), ] where (\operatorname{Cl}_2) is the Clausen function. The usual exam question uses the upper limit (\pi/2).

Theorem (Term-by-term differentiation of a power series)

Statement. Let the power series $\sum_{n=0}^{\infty} a_n x^n$ converge for $|x| < R$ (radius of convergence $R > 0$), and define

$f(x) = \sum_{n=0}^{\infty} a_n x^n, \qquad |x| < R.$

Then:

1. The series converges uniformly on every closed interval $[-R+\varepsilon,\; R-\varepsilon]$, for any $\varepsilon > 0$ chosen.
2. The function $f$ is continuous on $(-R, R)$.
3. The function $f$ is differentiable on $(-R, R)$, and its derivative is obtained by term-by-term differentiation:

$f'(x) = \sum_{n=1}^{\infty} n\,a_n x^{n-1}, \qquad |x| < R.$

Proof

Step 1: A useful bound on the coefficients

Since $\sum a_n x^n$ converges at some point $x_0 \neq 0$ with $|x_0| < R$, the terms $a_n x_0^{\,n}$ are bounded. So there exists $M > 0$ such that

$|a_n|\,|x_0|^n \le M \quad \text{for all } n.$

Hence

$|a_n| \le \frac{M}{|x_0|^n}.$

Step 2: Uniform convergence on $[-R+\varepsilon, R-\varepsilon]$

Fix $\varepsilon > 0$ and choose $r$ with $0 < r < R$, say $r = R - \varepsilon$ (taking $\varepsilon < R$; if $\varepsilon \ge R$ the interval is empty or trivial). For $|x| \le r$,

$|a_n x^n| \le |a_n| r^n \le M\left(\frac{r}{|x_0|}\right)^n.$

Choose $x_0$ with $|x_0| > r$ (possible since $r < R$). Then $\dfrac{r}{|x_0|} = \rho < 1$, so

$|a_n x^n| \le M\rho^n.$

Since $\sum M\rho^n$ is a convergent geometric series of constants (independent of $x$), the Weierstrass M-test gives that $\sum a_n x^n$ converges uniformly on $[-r, r] = [-R+\varepsilon, R-\varepsilon]$.

Step 3: Continuity of $f$

Each partial sum $S_N(x) = \sum_{n=0}^{N} a_n x^n$ is a polynomial, hence continuous on $(-R, R)$. By Step 2 the series converges uniformly on every closed subinterval of $(-R, R)$. A uniformly convergent sequence of continuous functions has a continuous limit. Therefore $f$ is continuous on $(-R, R)$.

Step 4: Differentiability and term-by-term differentiation

Consider the differentiated series

$g(x) = \sum_{n=1}^{\infty} n\,a_n x^{n-1}.$

We first show it has the same radius of convergence $R$. Using Step 1,

$|n\,a_n x^{n-1}| \le \frac{M\,n}{|x_0|^{\,n}}\,|x|^{n-1} = \frac{M}{|x|}\cdot n\left(\frac{|x|}{|x_0|}\right)^n.$

For $|x| < |x_0|$ the series $\sum n\left(\dfrac{|x|}{|x_0|}\right)^n$ converges (ratio test: $\dfrac{(n+1)\rho^{n+1}}{n\rho^{n}} \to \rho < 1$). Hence $\sum n a_n x^{n-1}$ converges absolutely for $|x| < |x_0|$, and since $|x_0|$ can be taken arbitrarily close to $R$, its radius of convergence is at least $R$. It cannot exceed $R$ (integration would give back $\sum a_n x^n$), so it is exactly $R$.

By the same M-test argument as in Step 2, the differentiated series converges uniformly on $[-R+\varepsilon, R-\varepsilon]$.

Now apply the theorem on term-by-term differentiation of uniformly convergent series of differentiable functions: if $S_N \to f$ pointwise on $(-R,R)$ and $S_N' \to g$ uniformly on each closed subinterval, then $f$ is differentiable and $f' = g$. Since $S_N'(x) = \sum_{n=1}^{N} n a_n x^{n-1}$, we obtain

$f'(x) = \lim_{N\to\infty} S_N'(x) = \sum_{n=1}^{\infty} n\,a_n x^{n-1}, \qquad |x| < R.$

This proves all three assertions. $\blacksquare$

Remarks

• The result holds at every interior point of the interval of convergence, so $f$ is differentiable (hence continuous) on the whole open interval $(-R, R)$.
• The theorem can be applied repeatedly: $f$ possesses derivatives of all orders on $(-R,R)$, and

$f^{(k)}(x) = \sum_{n=k}^{\infty} n(n-1)\cdots(n-k+1)\,a_n x^{n-k}.$

• Example. For $f(x) = \sum_{n=0}^{\infty} x^n = \dfrac{1}{1-x}$ with $R = 1$, term-by-term differentiation gives

$f'(x) = \sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2}, \qquad |x| < 1,$

which agrees with the derivative of $\dfrac{1}{1-x}$.

Let the given series be (\sum_{k=0}^{\infty} a_k), where
[ a_k=\frac{2^k x^{2^k-1}}{1+x^{2^k}},\qquad k=0,1,2,\dots ] For (|x|<1), define [ b_k=\frac{2^k x^{2^k-1}}{1-x^{2^k}}. ] Then [ \begin{aligned} b_k-b_{k+1} &=\frac{2^k x^{2^k-1}}{1-x^{2^k}} -\frac{2^{k+1}x^{2^{k+1}-1}}{1-x^{2^{k+1}}}\ &=\frac{2^k x^{2^k-1}}{1-x^{2^k}} -\frac{2^{k+1}x^{2^k-1}x^{2^k}}{(1-x^{2^k})(1+x^{2^k})}\ &=\frac{2^k x^{2^k-1}}{1-x^{2^k}} \left(1-\frac{2x^{2^k}}{1+x^{2^k}}\right)\ &=\frac{2^k x^{2^k-1}}{1-x^{2^k}}\cdot \frac{1-x^{2^k}}{1+x^{2^k}}\ &=\frac{2^k x^{2^k-1}}{1+x^{2^k}}=a_k. \end{aligned} ] Hence the series telescopes: [ \sum_{k=0}^{n} a_k =\sum_{k=0}^{n}(b_k-b_{k+1}) =b_0-b_{n+1}. ] Now [ b_0=\frac{1}{1-x}, ] and [ b_{n+1}=\frac{2^{n+1}x^{2^{n+1}-1}}{1-x^{2^{n+1}}}. ] Therefore [ \sum_{k=0}^{n} a_k =\frac{1}{1-x} -\frac{2^{n+1}x^{2^{n+1}-1}}{1-x^{2^{n+1}}}. ] It remains to show that the last term tends to (0) as (n\to\infty). For (x=0), this is trivial. For (0<|x|<1), put (m=2^{n+1}) and (a=-\ln|x|>0). Then [ m|x|^m=m e^{-am}\to 0\qquad (m\to\infty). ] Thus [ \frac{2^{n+1}x^{2^{n+1}-1}}{1-x^{2^{n+1}}} =\frac{m x^m}{x(1-x^m)}\to 0. ] Hence [ \sum_{k=0}^{\infty} a_k =\lim_{n\to\infty}\sum_{k=0}^{n} a_k =\frac{1}{1-x}. ] Therefore, for (|x|<1), [ \frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+\cdots =\frac{1}{1-x}. ] So the given series converges to (\boxed{\frac{1}{1-x}}).

Remark on the function. As printed, $f=\dfrac{x^2-y^2}{x^2+y^2}$ cannot satisfy the statement: along the $x$-axis $f(x,0)=1$ and along the $y$-axis $f(0,y)=-1$, so $\lim_{(x,y)\to(0,0)}f$ does not exist, $f$ is not continuous at $(0,0)$, and $f_x(0,0)=\lim_{h\to0}\frac{1-0}{h}$ does not exist. The standard problem has $f(x,y)=\dfrac{x^2y^2}{x^2+y^2}$ for $(x,y)\ne(0,0)$ and $f(0,0)=0$; it is solved below.

Partial derivatives at the origin.

$f_x(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to0}\frac{0-0}{h}=0,\qquad f_y(0,0)=0.$

Partial derivatives elsewhere. For $(x,y)\ne(0,0)$:

$f_x=\frac{2xy^2(x^2+y^2)-x^2y^2\cdot 2x}{(x^2+y^2)^2}=\frac{2xy^4}{(x^2+y^2)^2},\qquad f_y=\frac{2x^4y}{(x^2+y^2)^2}.$

Continuity of $f_x$ and $f_y$ at $(0,0)$. Put $x=r\cos\theta,\ y=r\sin\theta$:

$|f_x|=\frac{2r^5|\cos\theta|\sin^4\theta}{r^4}=2r|\cos\theta|\sin^4\theta\le 2r\to0\quad (r\to0).$

So $\lim_{(x,y)\to(0,0)}f_x=0=f_x(0,0)$. Similarly $|f_y|\le 2r\to0=f_y(0,0)$. Hence $f_x$ and $f_y$ are both continuous at $(0,0)$.

Differentiability. A function whose partial derivatives exist in a neighbourhood of a point and are continuous at that point is differentiable there. Hence $f$ is differentiable at $(0,0)$.

Direct check: with $\Delta f=f(h,k)-f(0,0)=\dfrac{h^2k^2}{h^2+k^2}$ and $f_x(0,0)=f_y(0,0)=0$,

$\frac{|\Delta f-0\cdot h-0\cdot k|}{\sqrt{h^2+k^2}}=\frac{h^2k^2}{(h^2+k^2)^{3/2}}\le\frac{(h^2+k^2)^2/4}{(h^2+k^2)^{3/2}}=\tfrac14\sqrt{h^2+k^2}\to0,$

so $f$ is differentiable at $(0,0)$ with $df=0$.

Jacobian. If $x,y,z$ are functions of $u,v,w$, the Jacobian of $(x,y,z)$ with respect to $(u,v,w)$ is the determinant

$J=\frac{\partial(x,y,z)}{\partial(u,v,w)}=\begin{vmatrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}&\frac{\partial x}{\partial w}\\[2pt]\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}&\frac{\partial y}{\partial w}\\[2pt]\frac{\partial z}{\partial u}&\frac{\partial z}{\partial v}&\frac{\partial z}{\partial w}\end{vmatrix}.$

It is the factor by which volume elements change under the transformation: $dx\,dy\,dz=|J|\,du\,dv\,dw$.

Solving for $x,y,z$.

$z=uvw,\qquad y=uv-z=uv(1-w),\qquad x=u-(y+z)=u-uv=u(1-v).$

Partial derivatives.

$x_u=1-v,\ x_v=-u,\ x_w=0;\qquad y_u=v(1-w),\ y_v=u(1-w),\ y_w=-uv;\qquad z_u=vw,\ z_v=uw,\ z_w=uv.$

Determinant. Expanding along the first row:

$J=(1-v)\big[u(1-w)\,uv+uv\cdot uw\big]+u\big[v(1-w)\,uv+uv\cdot vw\big]$

$=(1-v)\,u^2v\,[(1-w)+w]+u\cdot uv^2\,[(1-w)+w]=(1-v)u^2v+u^2v^2=u^2v.$

Hence $\dfrac{\partial(x,y,z)}{\partial(u,v,w)}=u^2v$.

Write $f(x,y)=\dfrac{x^2-y^2}{x^2+y^2}=1-\dfrac{2y^2}{x^2+y^2}=\dfrac{2x^2}{x^2+y^2}-1$.

Left side: integrate with respect to $y$ first. For fixed $x>0$:

$\int_0^1\left(\frac{2x^2}{x^2+y^2}-1\right)dy=2x\tan^{-1}\frac1x-1.$

Then

$I_1=\int_0^1\Big(2x\tan^{-1}\frac1x-1\Big)dx.$

Integrating by parts with $u=\tan^{-1}(1/x)$, $dv=2x\,dx$ (so $du=-\dfrac{dx}{1+x^2}$, $v=x^2$):

$\int_0^12x\tan^{-1}\frac1x\,dx=\Big[x^2\tan^{-1}\frac1x\Big]_0^1+\int_0^1\frac{x^2}{1+x^2}dx=\frac\pi4+\Big[x-\tan^{-1}x\Big]_0^1=\frac\pi4+1-\frac\pi4=1.$

So $I_1=1-1=0$.

Right side: integrate with respect to $x$ first. For fixed $y>0$:

$\int_0^1\Big(1-\frac{2y^2}{x^2+y^2}\Big)dx=1-2y\tan^{-1}\frac1y,$

$I_2=\int_0^1\Big(1-2y\tan^{-1}\frac1y\Big)dy=1-1=0,$

using the same integral as above (with $y$ in place of $x$).

Therefore

$\int_0^1dx\int_0^1\frac{x^2-y^2}{x^2+y^2}dy=\int_0^1dy\int_0^1\frac{x^2-y^2}{x^2+y^2}dx=0,$

and the two iterated integrals are equal. (The function is not continuous at the origin, so a double integral need not be defined by Fubini; here both iterated integrals happen to agree.)

Let (F_d(A)) denote the boundary (frontier) of (A) in the metric space ((X,d)). By definition,

[ F_d(A)=\bar A\cap \overline{X\setminus A}. ]

Equivalently, (x\in F_d(A)) iff every open ball (B(x,r)) intersects both (A) and (X\setminus A).

(i) Show that (F_d(A)=\bar A-\operatorname{Int}(A))

We use the standard identity

[ \operatorname{Int}(A)=X\setminus \overline{X\setminus A}. ]

Taking complements,

[ X\setminus \operatorname{Int}(A)=\overline{X\setminus A}. ]

Therefore,

[ F_d(A)=\bar A\cap \overline{X\setminus A} =\bar A\cap \bigl(X\setminus \operatorname{Int}(A)\bigr) =\bar A-\operatorname{Int}(A). ]

Hence,

[ \boxed{F_d(A)=\bar A-\operatorname{Int}(A)}. ]

(ii) Show that (F_d(A)=\varnothing) iff (A) is both open and closed in (X)

Forward part

Suppose

[ F_d(A)=\varnothing. ]

Using part (i),

[ \bar A-\operatorname{Int}(A)=\varnothing. ]

This means

[ \bar A\subseteq \operatorname{Int}(A). ]

But always,

[ \operatorname{Int}(A)\subseteq A\subseteq \bar A. ]

Hence,

[ \operatorname{Int}(A)=A=\bar A. ]

Now,

[ \operatorname{Int}(A)=A ]

means (A) is open, and

[ \bar A=A ]

means (A) is closed. Therefore (A) is both open and closed in (X).

Converse part

Conversely, suppose (A) is both open and closed in (X).

Since (A) is open,

[ \operatorname{Int}(A)=A. ]

Since (A) is closed,

[ \bar A=A. ]

Using part (i),

[ F_d(A)=\bar A-\operatorname{Int}(A) =A-A =\varnothing. ]

Thus,

[ \boxed{F_d(A)=\varnothing}. ]

Therefore,

[ \boxed{F_d(A)=\varnothing \iff A \text{ is both open and closed in } X.} ]

Hence both results are proved.

Part 1: relative error of $f=x^2y^2z^2$ at $x=y=z=1$.

Taking logarithms, $\log f=2\log x+2\log y+2\log z$, so

$\frac{df}{f}=2\left(\frac{dx}{x}+\frac{dy}{y}+\frac{dz}{z}\right).$

With $x=y=z=1$ and $dx=dy=dz=0.001$:

$\frac{\Delta f}{f}=2(0.001+0.001+0.001)=0.006,$

that is, a relative error of $0.006$ (about $0.6\%$). The absolute error is $\Delta f\approx0.006$, since $f(1,1,1)=1$.

OR — Part 2: root of $x^3-2x-5=0$ by bisection.

Let $f(x)=x^3-2x-5$. Then $f(2)=-1<0$ and $f(3)=16>0$, so a root lies in $(2,3)$.

The root lies in $(2.09375,\,2.09473)$, so to three decimal places

$x\approx 2.095.$

Uniform Limit of Continuous Functions is Continuous

Theorem. If $(f_n)$ is a sequence of continuous functions on $[a,b]$ and $f_n \to f$ uniformly on $[a,b]$, then $f$ is continuous on $[a,b]$.

Proof. Let $x_0 \in [a,b]$ be arbitrary. We show $f$ is continuous at $x_0$.

Let $\varepsilon > 0$. Since $f_n \to f$ uniformly, there exists $N \in \mathbb{N}$ such that for all $n \ge N$ and all $x \in [a,b]$, $|f_n(x) - f(x)| < \frac{\varepsilon}{3}.$

Fix such an $n \ge N$. Since $f_n$ is continuous at $x_0$, there exists $\delta > 0$ such that for all $x \in [a,b]$ with $|x - x_0| < \delta$, $|f_n(x) - f_n(x_0)| < \frac{\varepsilon}{3}.$

Now for any $x \in [a,b]$ with $|x - x_0| < \delta$, using the triangle inequality, $|f(x) - f(x_0)| \le |f(x) - f_n(x)| + |f_n(x) - f_n(x_0)| + |f_n(x_0) - f(x_0)|$ $< \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon.$

Hence $f$ is continuous at $x_0$. Since $x_0 \in [a,b]$ was arbitrary, $f$ is continuous on $[a,b]$. $\blacksquare$

Pointwise Convergence of $f_n(x) = x^n$ on $[0,1]$

For each $n$, $f_n(x) = x^n$ is continuous on $[0,1]$. We find the pointwise limit.

• If $0 \le x < 1$, then $x^n \to 0$ as $n \to \infty$.
• If $x = 1$, then $x^n = 1$ for all $n$, so $x^n \to 1$.

Therefore the pointwise limit is $f(x) = \lim_{n\to\infty} x^n = \begin{cases} 0, & 0 \le x < 1,\\[4pt] 1, & x = 1. \end{cases}$

So $(f_n)$ converges pointwise to $f$ on $[0,1]$.

The Convergence is Not Uniform

Suppose, for contradiction, that $f_n \to f$ uniformly on $[0,1]$. Since each $f_n$ is continuous on $[0,1]$, by the theorem proved above the limit function $f$ must be continuous on $[0,1]$.

But the limit function found above, $f(x) = \begin{cases} 0, & 0 \le x < 1,\\[4pt] 1, & x = 1, \end{cases}$ is not continuous at $x = 1$: indeed, taking $x_n = 1 - \tfrac{1}{n} \to 1$, we get $f(x_n) = 0 \to 0$, while $f(1) = 1$. Thus $\lim_{x \to 1^-} f(x) = 0 \ne 1 = f(1)$.

This contradiction shows that the convergence cannot be uniform.

Direct verification. For uniform convergence we would need $\sup_{x \in [0,1]} |f_n(x) - f(x)| \to 0.$ But for each $n$, choosing $x = \left(\tfrac{1}{2}\right)^{1/n} \in [0,1)$ gives $|f_n(x) - f(x)| = \left|\left(\left(\tfrac12\right)^{1/n}\right)^n - 0\right| = \frac12.$ Hence $\sup_{x \in [0,1]} |f_n(x) - f(x)| \ge \dfrac12$ for every $n$, so the supremum does not tend to $0$. Therefore the convergence is not uniform. $\blacksquare$

Solution (first alternative):

Let
[ d(F_n)=\operatorname{diam}(F_n)=\sup{d(x,y):x,y\in F_n}. ] Given that (d(F_n)\to 0) as (n\to\infty).

For each (n), choose a point (x_n\in F_n), which is possible because each (F_n) is non-empty.

1. The sequence ((x_n)) is Cauchy

Let (\varepsilon>0). Since (d(F_n)\to 0), there exists (N\in\mathbb N) such that [ d(F_n)<\varepsilon \qquad \text{for all } n\ge N. ] Now take (m,n\ge N). Since the sets are nested, [ F_m\subseteq F_N,\qquad F_n\subseteq F_N. ] Hence (x_m,x_n\in F_N). Therefore, [ d(x_m,x_n)\le d(F_N)<\varepsilon. ] Thus ((x_n)) is a Cauchy sequence in (X).

2. The intersection is non-empty

Since ((X,d)) is complete, the Cauchy sequence ((x_n)) converges to some point (x\in X), say [ x_n\to x. ]

Fix any (k\in\mathbb N). For all (n\ge k), we have [ x_n\in F_n\subseteq F_k. ] So the tail of the sequence ((x_n)) lies in (F_k). Since (F_k) is closed, its limit (x) must also lie in (F_k). Thus [ x\in F_k \qquad \text{for every } k\in\mathbb N. ] Therefore, [ x\in \bigcap_{n=1}^{\infty}F_n. ] So the intersection is non-empty.

3. The intersection contains exactly one point

Suppose (x,y\in \bigcap_{n=1}^{\infty}F_n). Then for every (n), [ x,y\in F_n. ] Hence [ d(x,y)\le d(F_n) \qquad \text{for all } n. ] But (d(F_n)\to 0). Therefore, [ d(x,y)=0. ] By the metric axioms, (d(x,y)=0) implies (x=y).

Thus the intersection cannot contain two distinct points. Hence [ \bigcap_{n=1}^{\infty}F_n ] contains exactly one point. (\square)