1.5: Dirichlet’s Tests

Weierstrass’s test is useful and important, but it has a basic shortcoming: it applies only to absolutely uniformly convergent improper integrals. The next theorem applies in some cases where $\int_{a}^{b}f(x,y)\,dx$ converges uniformly on $S$, but $\int_{a}^{b}|f(x,y)|\,dx$ does not.

[theorem:8] (Dirichlet’s Test for Uniform Convergence I) If $g,$ $g_{x},$ and $h$ are continuous on $[a,b)\times S,$ then

$$\int_{a}^{b}g(x,y)h(x,y)\,dx$$

converges uniformly on $S$ if the following conditions are satisfied:

$$\displaystyle{\lim_{x\to b-}\left\{\sup_{y\in S}|g(x,y)|\right\}=0};$$

There is a constant $M$ such that

$$\sup_{y\in S}\left|\int_{a}^{x}h(u,y)\,du\right|< M, \quad a\le x<b;$$

$\int_{a}^{b}|g_{x}(x,y)|\,dx$ converges uniformly on $S.$

If

$$\label{eq:20} H(x,y)=\int_{a}^{x}h(u,y)\,du,$$

then integration by parts yields

$$\begin{aligned} \int_{r}^{r_{1}}g(x,y)h(x,y)\,dx&=&\int_{r}^{r_{1}}g(x,y)H_{x}(x,y)\,dx \nonumber\\ &=&g(r_{1},y)H(r_{1},y)-g(r,y)H(r,y)\label{eq:21}\\ &&-\int_{r}^{r_{1}}g_{x}(x,y)H(x,y)\,dx. \nonumber\end{aligned}$$

Since assumption (b) and [eq:20] imply that $|H(x,y)|\le M,$ $(x,y)\in (a,b]\times S$, Eqn. [eq:21] implies that

$$\label{eq:22} \left|\int_{r}^{r_{1}}g(x,y)h(x,y)\,dx\right|< M\left(2\sup_{x\ge r}|g(x,y)|+\int_{r}^{r_{1}}|g_{x}(x,y)|\,dx\right)$$

on $[r,r_{1}]\times S$.

Now suppose $\epsilon>0$. From assumption (a), there is an $r_{0} \in [a,b)$ such that $|g(x,y)|<\epsilon$ on $S$ if $r_{0}\le x <b$. From assumption (c) and Theorem [theorem:6], there is an $s_{0}\in [a,b)$ such that

$$\int_{r}^{r_{1}}|g_{x}(x,y)|\,dx<\epsilon, \quad y\in S, \quad s_{0}<r<r_{1}<b.$$

Therefore [eq:22] implies that

$$\left|\int_{r}^{r_{1}}g(x,y)h(x,y)\right| < 3M\epsilon, \quad y\in S, \quad \max(r_{0},s_{0})<r<r_{1}<b.$$

Now Theorem [theorem:4] implies the stated conclusion.

The statement of this theorem is complicated, but applying it isn’t; just look for a factorization $f=gh$, where $h$ has a bounded antderivative on $[a,b)$ and $g$ is “small” near $b$. Then integrate by parts and hope that something nice happens. A similar comment applies to Theorem 9, which follows.

Example $\PageIndex{1}$

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Solution

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Let

$$I(y)=\int_{0}^{\infty}\frac{\cos xy}{x+y}\,dx,\quad y>0.$$

The obvious inequality

$$\left|\frac{\cos xy}{x+y}\right|\le \frac{1}{x+y}$$

is useless here, since

$$\int_{0}^{\infty}\frac{dx}{x+y}=\infty.$$

However, integration by parts yields

$$\begin{aligned} \int_{r}^{r_{1}}\frac{\cos xy}{x+y}\,dx &=& \frac{\sin xy}{y(x+y)}\biggr|_{r}^{r_{1}}+ \int_{r}^{r_{1}}\frac{\sin xy}{y(x+y)^{2}}\,dx\\ &=&\frac{\sin r_{1}y}{y(r_{1}+y)}-\frac{\sin ry}{y(r+y)} +\int_{r}^{r_{1}}\frac{\sin xy}{y(x+y)^{2}}\,dx.\end{aligned}$$

Therefore, if $0< r<r_{1}$, then

$$\begin{aligned} \left|\int_{r}^{r_{1}}\frac{\cos xy}{x+y}\,dx\right|< \frac{1}{y}\left(\frac{2}{r+y}+\int_{r}^{\infty}\frac{1}{(x+y)^{2}}\right) \le \frac{3}{y(r+y)^{2}}\le \frac{3}{\rho(r+\rho)}\end{aligned}$$

if $y\ge \rho>0$. Now Theorem [theorem:4] implies that $I(y)$ converges uniformly on $[\rho,\infty)$ if $\rho>0$.

We leave the proof of the following theorem to you (Exercise [exer:10]).

[theorem:9] (Dirichlet’s Test for Uniform Convergence II) If $g,$ $g_{x},$ and $h$ are continuous on $(a,b]\times S,$ then

$$\int_{a}^{b}g(x,y)h(x,y)\,dx$$

converges uniformly on $S$ if the following conditions are satisfied:

$\displaystyle{\lim_{x\to a+}\left\{\sup_{y\in S}|g(x,y)|\right\}=0};$

There is a constant $M$ such that

$$\sup_{y\in S}\left|\int_{x}^{b}h(u,y)\,du\right| \le M, \quad a< x\le b;$$

$\int_{a}^{b}|g_{x}(x,y)|\,dx$ converges uniformly on $S$.

By recalling Theorems 3.4.10 (p. 163), 4.3.20 (p. 217), and 4.4.16 (p. 248), you can see why we associate Theorems [theorem:8] and [theorem:9] with Dirichlet.