1.4: Absolutely Uniformly Convergent Improper Integrals

(Absolute Uniform Convergence I) [definition:4] The improper integral

\[\int_{a}^{b}f(x,y)\,dx=\lim_{r\to b-}\int_{a}^{r}f(x,y)\,dx\]

is said to converge absolutely uniformly on \(S\) if the improper integral

\[\int_{a}^{b}|f(x,y)|\,dx=\lim_{r\to b-}\int_{a}^{r}|f(x,y)|\,dx\]

converges uniformly on \(S\); that is, if, for each \(\epsilon>0\), there is an \(r_{0}\in [a,b)\) such that

\[\left|\int_{a}^{b}|f(x,y)|\,dx-\int_{a}^{r}|f(x,y)|\,dx\right| <\epsilon, \quad y\in S,\quad r_{0}<r<b.\]

To see that this definition makes sense, recall that if \(f\) is locally integrable on \([a,b)\) for all \(y\) in \(S\), then so is \(|f|\) (Theorem 3.4.9, p. 161). Theorem [theorem:4] with \(f\) replaced by \(|f|\) implies that \(\int_{a}^{b}f(x,y)\,dx\) converges absolutely uniformly on \(S\) if and only if, for each \(\epsilon>0\), there is an \(r_{0}\in [a,b)\) such that

\[\int_{r}^{r_{1}}|f(x,y)|\,dx<\epsilon,\quad y\in S,\quad r_{0}\le r<r_{1}<b .\]

Since

\[\left|\int_{r}^{r_{1}}f(x,y)\,dx\right| \le \int_{r}^{r_{1}}|f(x,y)|\,dx,\]

Theorem [theorem:4] implies that if \(\int_{a}^{b}f(x,y)\,dx\) converges absolutely uniformly on \(S\) then it converges uniformly on \(S\).

[theorem:6] ( Weierstrass’s Test for Absolute Uniform Convergence I) Suppose \(M=M(x)\) is nonnegative on \([a,b),\) \(\int_{a}^{b}M(x)\,dx<\infty,\) and

\[\label{eq:19} |f(x,y)| \le M(x), \quad y\in S,\quad a\le x<b.\]

Then \(\int_{a}^{b}f(x,y)\,dx\) converges absolutely uniformly on \(S.\)

Denote \(\int_{a}^{b}M(x)\,dx=L<\infty\). By definition, for each \(\epsilon>0\) there is an \(r_{0}\in [a,b)\) such that

\[L-\epsilon < \int_{a}^{r}M(x)\,dx \le L,\quad r_{0}<r<b.\]

Therefore, if \(r_{0}< r\le r_{1},\) then

\[0\le \int_{r}^{r_{1}}M(x)\,dx=\left(\int_{a}^{r_{1}}M(x)\,dx -L\right)- \left(\int_{a}^{r}M(x)\,dx -L\right)<\epsilon\]

This and [eq:19] imply that

\[\int_{r}^{r_{1}}|f(x,y)|\,dx\le \int_{r}^{r_{1}} M(x)\,dx <\epsilon,\quad y\in S, \quad a\le r_{0}<r<r_{1}<b.\]

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

[example:7] Suppose \(g=g(x,y)\) is locally integrable on \([0,\infty)\) for all \(y\in S\) and, for some \(a_{0}\ge 0\), there are constants \(K\) and \(p_{0}\) such that

\[|g(x,y)| \le Ke^{p_{0}x},\quad y\in S, \quad x\ge a_{0}.\]

If \(p>p_{0}\) and \(r\ge a_{0}\), then

\[\begin{aligned} \int_{r}^{\infty}e^{-px} |g(x,y)|\,dx &=& \int_{r}^{\infty} e^{-(p-p_{0})x}e^{-p_{0}x}|g(x,y)|\,dx\\ &\le& K\int_{r}^{\infty} e^{-(p-p_{0})x}\,dx= \frac{K e^{-(p-p_{0})r}}{p-p_{0}},\end{aligned}\]

so \(\int_{0}^{\infty}e^{-px} g(x,y)\,dx\) converges absolutely on \(S\). For example, since

\[|x^{\alpha}\sin xy|<e^{p_{0}x}\text{\quad and \quad} |x^{\alpha}\cos xy|<e^{p_{0}x}\]

for \(x\) sufficiently large if \(p_{0}>0\), Theorem [theorem:4] implies that \(\int_{0}^{\infty}e^{-px}x^{\alpha}\sin xy\,dx\) and \(\int_{0}^{\infty}e^{-px}x^{\alpha}\cos xy\,dx\) converge absolutely uniformly on \((-\infty,\infty)\) if \(p>0\) and \(\alpha~\ge~0\). As a matter of fact, \(\int_{0}^{\infty}e^{-px}x^{\alpha}\sin xy\,dx\) converges absolutely on \((-\infty,\infty)\) if \(p>0\) and \(\alpha>-1\). (Why?)

(Absolute Uniform Convergence II) [definition:5] The improper integral

\[\int_{a}^{b}f(x,y)\,dx=\lim_{r\to a+}\int_{r}^{b}f(x,y)\,dx\]

is said to converge absolutely uniformly on \(S\) if the improper integral

\[\int_{a}^{b}|f(x,y)|\,dx=\lim_{r\to a+}\int_{r}^{b}|f(x,y)|\,dx\]

converges uniformly on \(S\); that is, if, for each \(\epsilon>0\), there is an \(r_{0}\in (a,b]\) such that

\[\left|\int_{a}^{b}|f(x,y)|\,dx-\int_{r}^{b}|f(x,y)|\,dx\right| <\epsilon, \quad y\in S, \quad a<r<r_{0}\le b.\]

We leave it to you (Exercise [exer:7]) to prove the following theorem.

[theorem:7] (Weierstrass’s Test for Absolute Uniform Convergence II) Suppose \(M=M(x)\) is nonnegative on \((a,b],\) \(\int_{a}^{b}M(x)\,dx<\infty,\) and

\[|f(x,y)| \le M(x), \quad y\in S, \quad x\in (a,b].\]

Then \(\int_{a}^{b}f(x,y)\,dx\) converges absolutely uniformly on \(S\).

[example:8] If \(g=g(x,y)\) is locally integrable on \((0,1]\) for all \(y\in S\) and

\[|g(x,y)| \le Ax^{-\beta}, \quad 0<x \le x_{0},\]

for each \(y \in S\), then

\[\int_{0}^{1} x^{\alpha}g(x,y)\,dx\]

converges absolutely uniformly on \(S\) if \(\alpha>\beta-1\). To see this, note that if \(0<r< r_{1}\le x_{0}\), then

\[\int_{r_{1}}^{r}x^{\alpha}|g(x,y)|\,dx \le A\int_{r_{1}}^{r} x^{\alpha-\beta}\,dx= \frac{Ax^{\alpha-\beta+1}}{\alpha-\beta+1}\biggr|_{r_{1}}^{r}< \frac{Ar^{\alpha-\beta+1}}{\alpha-\beta+1}.\]

Applying this with \(\beta=0\) shows that

\[F(y)=\int_{0}^{1} x^{\alpha}\cos xy\,dx\]

converges absolutely uniformly on \((-\infty,\infty)\) if \(\alpha>-1\) and

\[G(y)=\int_{0}^{1}x^{\alpha}\sin xy \,dx\]

converges absolutely uniformly on \((-\infty,\infty)\) if \(\alpha>-2\).

By recalling Theorem 4.4.15 (p. 246), you can see why we associate Theorems [theorem:6] and [theorem:7] with Weierstrass.