Exercise 6.9

Question

Exercise 9: Show that integration by parts can sometimes be applied to the ``improper'' integrals defined in Exercises 7 and 8. (State appropriate hypotheses, formulate a theorem,
    and prove it.) For instance, show that $$\int_0^{\infty}\frac{\cos x}{1+x}\,dx=\int_0^{\infty}\frac{\sin x}{(1+x)^2}\,dx.$$ Show that one of these integrals converges {\it absolutely,}
    but that the other does not.

analambanomenos · Accepted Answer

If $F$ and $G$ are functions which are differentiable on $[c,1]$ for all $c>0$ and such that $F'=f\in\mathscr R$ and $G'=g\in\mathscr R$ on $[c,1]$ for all $c>0$, then by Theorem
    6.22 we have $$\int_c^1F(x)g(x)\,dx=F(1)G(1)-F(c)G(c)-\int_c^1f(x)G(x)\,dx.$$ Suppose that two of the limits $$\lim_{cightarrow0}\int_c^1F(x)g(x)\,dx=\int_0^1F(x)g(x)\,dx,\;\;
    \lim_{cightarrow0}\int_c^1f(x)G(x)\,dx=\int_0^1f(x)G(x)\,dx,\;\;\lim_{cightarrow0}F(c)G(c)$$ exist and are finite. Then the third limit exists and is finite, and we have
    $$\int_0^1F(x)g(x)\,dx=F(1)G(1)-\lim_{cightarrow0}F(c)G(c)-\int_0^1F(x)g(x)\,dx.$$ Similarly, if $F$ and $G$ are functions which are differentiable on $[a,b]$ for all $b>a$ and
    such that $F'=f\in\mathscr R$ and $G'=g\in\mathscr R$ on $[a,b]$ for all $b>a$, then by Theorem 6.22 we have $$\int_a^bF(x)g(x)\,dx=F(b)G(b)-F(a)G(a)-\int_a^bf(x)G(x)\,dx.$$
    Suppose that two of the limits $$\lim_{bightarrow\infty}\int_a^bF(x)g(x)\,dx=\int_a^{\infty} F(x)g(x)\,dx,\;\;\lim_{bightarrow\infty}\int_a^bf(x)G(x)\,dx=\int_a^{\infty}
    f(x)G(x)\,dx,\;\;\lim_{bightarrow\infty}F(b)G(b)$$ exist and are finite. Then the third limit exists and is finite, and we have $$\int_a^{\infty} F(x)g(x)\,dx=
    \lim_{bightarrow\infty}F(b)G(b)-F(a)G(a)-\int_a^{\infty} F(x)g(x)\,dx.$$

For the example, let $$F(x)=\frac{1}{1+x},\quad F'(x)=f(x)=-\frac{1}{(1+x)^2},\quad G(x)=\sin x,\quad G'(x)=g(x)=\cos x.$$ The functions $F$ and $G$ are differentiable on $[0,b)$,
    for all $b>0$, and $f\in\mathscr R$, $g\in\mathscr R$ on $[0,b]$ for all $b>0$. Also $$\lim_{bightarrow\infty}F(b)G(b)=\lim_{bightarrow\infty}\frac{\sin b}{1+b}=0$$ and
    $$\lim_{bightarrow\infty}\bigg|\int_0^b\frac{\sin x}{(1+x)^2}\,dx\bigg|\le\lim_{bightarrow\infty}\int_0^b\frac{|\sin x|}{(1+x)^2}\,dx\le\lim_{bightarrow\infty}
    \int_0^b\frac{1}{(1+x)^2}\,dx$$ which converges by Exercise 8 since $\sum_0^{\infty} 1/(1+n)^2$ converges. Hence we can apply the results of the first part of this exercise and 
    conclude that $$\int_0^{\infty}\frac{\cos x}{1+x}\,dx=\lim_{bightarrow\infty}\frac{\sin b}{1+b}-\frac{\sin 0}{1+0}+\int_0^{\infty}\frac{\sin x}{(1+x)^2}\,dx=\int_0^{\infty}
    \frac{\sin x}{(1+x)^2}\,dx.$$

We've seen above that $\sin x/(1+x)^2$ converges absolutely on $[0,\infty)$. To show that $\cos x/(1+x)$ diverges absolutely, 
    \begin{align*}
        \int_0^{\infty}\frac{|\cos x|}{1+x}\,dx &= \sum_{k=0}^{\infty}\int_{2\pi k}^{2\pi(k+1)}\frac{|\cos x|}{1+x}\,dx \
        &\ge\sum_{k=0}^{\infty}\frac{1}{2\pi(k+1)+1}\int_{2\pi k}^{2\pi(k+1)}|\cos x|\,dx \
        &\ge\sum_{k=0}^{\infty}\frac{1}{2\pi(k+1)+2\pi}\int_0^{2\pi}|\cos x|\,dx \
        &=\frac{2}{\pi}\sum_{k=0}^{\infty}\frac{1}{k+2}
    \end{align*}
    a sum which diverges.

Exercise 6.9

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Exercise 6.9

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