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Mathematics LibreTexts

1.1: Introduction to Improper Functions

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    In Section 7.2 (pp. 462–484) we considered functions of the form

    \[F(y)=\int_{a}^{b}f(x,y)\,dx, \quad c \le y \le d.\]

    We saw that if \(f\) is continuous on \([a,b]\times [c,d]\), then \(F\) is continuous on \([c,d]\) (Exercise 7.2.3, p. 481) and that we can reverse the order of integration in


    to evaluate it as


    (Corollary 7.2.3, p. 466).

    Here is another important property of \(F\).

    [theorem:1] If \(f\) and \(f_{y}\) are continuous on \([a,b]\times [c,d],\) then

    \[\label{eq:1} F(y)=\int_{a}^{b}f(x,y)\,dx, \quad c \le y \le d,\]

    is continuously differentiable on \([c,d]\) and \(F'(y)\) can be obtained by differentiating [eq:1] under the integral sign with respect to \(y;\) that is,

    \[\label{eq:2} F'(y)=\int_{a}^{b}f_{y}(x,y)\,dx, \quad c \le y \le d.\]

    Here \(F'(a)\) and \(f_{y}(x,a)\) are derivatives from the right and \(F'(b)\) and \(f_{y}(x,b)\) are derivatives from the left\(.\)

    If \(y\) and \(y+\Delta y\) are in \([c,d]\) and \(\Delta y\ne0\), then

    \[\label{eq:3} \frac{F(y+\Delta y)-F(y)}{\Delta y}= \int_{a}^{b}\frac{f(x,y+\Delta y)-f(x,y)}{\Delta y}\,dx.\]

    From the mean value theorem (Theorem 2.3.11, p. 83), if \(x\in[a,b]\) and \(y\), \(y+\Delta y\in[c,d]\), there is a \(y(x)\) between \(y\) and \(y+\Delta y\) such that

    \[f(x,y+\Delta y)-f(x,y)=f_{y}(x,y)\Delta y= f_{y}(x,y(x))\Delta y+(f_{y}(x,y(x)-f_{y}(x,y))\Delta y.\]

    From this and [eq:3],

    \[\label{eq:4} \left|\frac{F(y+\Delta y)-F(y)}{\Delta y}-\int_{a}^{b}f_{y}(x,y)\,dx\right| \le \int_{a}^{b} |f_{y}(x,y(x))-f_{y}(x,y)|\,dx.\]

    Now suppose \(\epsilon>0\). Since \(f_{y}\) is uniformly continuous on the compact set \([a,b]\times [c,d]\) (Corollary 5.2.14, p. 314) and \(y(x)\) is between \(y\) and \(y+\Delta y\), there is a \(\delta>0\) such that if \(|\Delta|<\delta\) then

    \[|f_{y}(x,y)-f_{y}(x,y(x))|<\epsilon,\quad (x,y)\in[a,b]\times [c,d].\]

    This and [eq:4] imply that

    \[\left|\frac{F(y+\Delta y-F(y))}{\Delta y}-\int_{a}^{b}f_{y}(x,y)\,dx\right|<\epsilon(b-a)\]

    if \(y\) and \(y+\Delta y\) are in \([c,d]\) and \(0<|\Delta y|<\delta\). This implies [eq:2]. Since the integral in [eq:2] is continuous on \([c,d]\) (Exercise 7.2.3, p. 481, with \(f\) replaced by \(f_{y}\)), \(F'\) is continuous on \([c,d]\).

    [example:1] Since

    \[f(x,y)=\cos xy\text{\quad and\quad} f_{y}(x,y)=-x\sin xy\]

    are continuous for all \((x,y)\), Theorem [theorem:1] implies that if

    \[\label{eq:5} F(y)=\int_{0}^{\pi} \cos xy\,dx,\quad -\infty<y<\infty,\]


    \[\label{eq:6} F'(y)=-\int_{0}^{\pi}x\sin xy\,dx,\quad -\infty<y<\infty.\]

    (In applying Theorem [theorem:1] for a specific value of \(y\), we take \(R=[0,\pi]\times [-\rho,\rho]\), where \(\rho>|y|\).) This provides a convenient way to evaluate the integral in [eq:6]: integrating the right side of [eq:5] with respect to \(x\) yields

    \[F(y)=\frac{\sin xy}{y}\bigg|_{x=0}^{\pi}=\frac{\sin\pi y}{y}, \quad y\ne0.\]

    Differentiating this and using [eq:6] yields

    \[\int_{0}^{\pi}x\sin xy\,dx =\frac{\sin \pi y}{y^{2}}- \frac{\pi\cos \pi y}{y}, \quad y\ne0.\]

    To verify this, use integration by parts.

    We will study the continuity, differentiability, and integrability of

    \[F(y)=\int_{a}^{b}f(x,y)\,dx,\quad y\in S,\]

    where \(S\) is an interval or a union of intervals, and \(F\) is a convergent improper integral for each \(y\in S\). If the domain of \(f\) is \([a,b)\times S\) where \(-\infty<a< b\le \infty\), we say that \(F\) is pointwise convergent on \(S\) or simply convergent on \(S\), and write

    \[\label{eq:7} \int_{a}^{b}f(x,y)\,dx=\lim_{r\to b-}\int_{a}^{r}f(x,y)\,dx\]

    if, for each \(y\in S\) and every \(\epsilon>0\), there is an \(r=r_{0}(y)\) (which also depends on \(\epsilon\)) such that

    \[\label{eq:8} \left|F(y)-\int_{a}^{r}f(x,y)\,dx\right|= \left|\int_{r}^{b}f(x,y)\,dx\right|< \epsilon, \quad r_{0}(y)\le y<b.\]

    If the domain of \(f\) is \((a,b]\times S\) where \(-\infty\le a<b<\infty\), we replace [eq:7] by

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

    and [eq:8] by

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

    In general, pointwise convergence of \(F\) for all \(y\in S\) does not imply that \(F\) is continuous or integrable on \([c,d]\), and the additional assumptions that \(f_{y}\) is continuous and \(\int_{a}^{b}f_{y}(x,y)\,dx\) converges do not imply [eq:2].

    [example:2] The function


    is continuous on \([0,\infty)\times (-\infty,\infty)\) and

    \[F(y)=\int_{0}^{\infty}f(x,y)\,dx =\int_{0}^{\infty}ye^{-|y|x}\,dx\]

    converges for all \(y\), with

    \[F(y)= \begin{cases} -1& y<0,\\ \phantom{-}0&y=0,\\ \phantom{-}1&y>0;\\ \end{cases}\]

    therefore, \(F\) is discontinuous at \(y=0\).

    [example:3] The function


    is continuous on \([0,\infty)\times (-\infty,\infty)\). Let

    \[F(y)=\int_{0}^{\infty}f(x,y)\,dx= \int_{0}^{\infty}y^{3}e^{-y^{2}x}\,dx =y,\quad -\infty<y<\infty.\]


    \[F'(y)=1, \quad -\infty<y<\infty.\]


    \[\int_{0}^{\infty}\frac{\partial{}}{\partial{y}}(y^{3}e^{-y^{2}x})\,dx =\int_{0}^{\infty}(3y^{2}-2y^{4}x)e^{-y^{2}x}\,dx= \begin{cases} 1,& y\ne0,\\ 0,& y=0, \end{cases}\]


    \[F'(y)\ne\int_{0}^{\infty}\frac{\partial{f(x,y)}}{\partial{y}}\,dx\text{\quad if\quad}y=0.\]