Definition $10.0 .1$ The term $L^{1}(a, b)$ where $a \geq-\infty$ and $b \leq \infty$ will refer to functions which are Riemann integrable on finite intervals properly contained in $(a, b)$ and $\lim {r \rightarrow a+, R \rightarrow b-} \int{r}^{R}|f(t)| d t$ exists and the value of this limit will be denoted by $\int_{a}^{b}|f(t)| d t$. The symbol means that whenever $\varepsilon>0$, there exists $\tilde{r}>a$ and $\tilde{R}\tilde{R}$, you have the number denoted as $\int_{a}^{b}|f(t)| d t$ satisfying
$$
\left|\int_{a}^{b}\right| f(t)\left|d t-\int_{r}^{R}\right| f(t)|d t|<\varepsilon
$$
(When you have a decent integral, this whole presentation is a lot easier.) The claims of the following example are easy and are left for you to verify.
Example 10.0.2 If $f(x)=x^{-p}$, for $p \in(0,1)$, then $f \in L^{1}(0,1)$. If $f(x)=e^{-x}$, then $f \in L^{1}(-a, \infty)$ for any $a<\infty$. If $f(x)=1 / x^{2}$, then $f \in L^{1}(1, \infty)$. If $f(x)=e^{-t} t^{-p}$ where $p>0$, then $f \in L^{1}(0, \infty)$.
Lemma 10.0.3 Suppose $f \in L^{1}(a, b)$. Then for $g(t)$ any bounded continuous function, say $|g(t)| \leq M$
$$
\lim {r \rightarrow a+, R \rightarrow b-} \int{r}^{R} f(t) g(t) d t \text { exists }
$$
and this will be denoted as $\int_{a}^{b} f(t) g(t) d t$. It is called the improper integral. Also, it can be computed by letting $r_{n}$ be a decreasing sequence converging to a and $R_{n}$ an increasing sequence converging to $b$ and then
$$
\int_{a}^{b} f(t) g(t) d t=\lim {n \rightarrow \infty} \int{r_{n}}^{R_{n}} f(t) g(t) d t
$$
Proof: There is no issue about the existence of $\int_{r}^{R} f(t) g(t) d t$ thanks to Theorem $9.3 .12$. Let $\tilde{R}, \tilde{r}$ be as described in the definition corresponding to $\frac{\varepsilon}{5 M}$ for $|f(t)|$ in place of $f(t)$. Then let $r, \hat{r}<\tilde{r}, R, \hat{R}>\tilde{R}$. Then $\left|\int_{r}^{R} f(t) g(t) d t-\int_{\hat{r}}^{\hat{R}} f(t) g(t) d t\right| \leq \int_{\min (r, \hat{r})}^{\max (r, \hat{r})}|f(t) g(t)| d t+\int_{\min (R, \hat{R})}^{\max (R, \hat{R})}|f(t) g(t)| d t$ 219 CHAPTER 10. IMPROPER INTEGRALS
$$
\begin{aligned}
&\leq M \int_{\min (r, \hat{r})}^{\bar{r}}|f(t)| d t+M \int_{\bar{R}}^{\max (R, \hat{R})}|f(t)| d t \
&\leq M\left(\int_{r}^{\bar{r}}|f(t)| d t+\int_{\bar{R}}^{R}|f(t)| d t+\int_{\hat{r}}^{\bar{r}}|f(t)| d t+\int_{\bar{R}}^{\hat{R}}|f(t)| d t\right)<4 M \frac{\varepsilon}{5 M}=\frac{4 \varepsilon}{5} \end{aligned} $$ Thus, letting $\varepsilon_{n}$ decrease to 0 , there exists $\tilde{r}_{n}, \tilde{R}_{n}$ with $\tilde{r}_{n}$ decreasing to $a$ and $\tilde{R}_{n}$ increasing to $b$ such that the diameters of the set $$ S_{n} \equiv \overline{\left\{\int_{r}^{R} f(t) g(t) d t: aR>\tilde{R}{n}\right}} $$ is no more than $\frac{4}{5} \varepsilon{n}$. By completeness, there is $I \in \cap_{n} S_{n}$. Then if $r<\tilde{r}{n}, R>\tilde{R}{n}$, $$ \left|I-\int_{r}^{R} f(t) g(t) d t\right| \leq \frac{4}{5} \varepsilon_{n}<\varepsilon_{n} $$ and so the improper integral exists and equals $I$. Let $\varepsilon>0$ be given and let $\varepsilon_{n}
<\varepsilon .$ Let $r_{k}$ decrease to $a$ and $R_{k}$ increase to
$b$, then for $k$ large enough, $r_{k}<\tilde{r}{n}$ and $\bar{R} {n}<R_{k}$ so $\int_{r_{k}}^{R_{k}} f(t) g(t) d t \in S_{n}$ and hence
$\left|I-\int_{r_{k}}^{R_{k}} f(t) g(t) d t\right| \leq \frac{4 \varepsilon}
{5}<\varepsilon$ which shows that
$$
\begin{aligned}
&=e(\varepsilon)+E(R)+\int_{0}^{R} \frac{1}{1+t^{2}} d t-\int_{0}^{R} e^{-R t} \frac{(\cos R)+t(\sin R)}{1+t^{2}} d t \
&=e(\varepsilon)+E(R)+\int_{0}^{R} \frac{1}{1+t^{2}} d t-\int_{0}^{R} e^{-R t} \frac{1}{\sqrt{1+t^{2}}} \cos \left(R-\phi_{t}\right) d t
\end{aligned}
$$
where $\phi_{t}$ is a phase shift. This is because
$$
\left(\frac{1}{\sqrt{1+t^{2}}}, \frac{t}{\sqrt{1+t^{2}}}\right)=\left(\cos \phi_{t}, \sin \phi_{t}\right) \text { some } \phi_{t}
$$
so the formula for $\cos (R-\phi)$ is used. This last integral satisfies
$$
\left|\int_{0}^{R} e^{-R t} \frac{1}{\sqrt{1+t^{2}}} \cos \left(R-\phi_{t}\right) d t\right| \leq \int_{0}^{R} e^{-R t} d t \leq \frac{1}{R}
$$
Therefore, include it in $E(R)$ and it follows that
$$
\int_{\varepsilon}^{R} \frac{\sin x}{x} d x=e(\varepsilon)+E(R)+\int_{0}^{R} \frac{1}{1+t^{2}} d t
$$
where $\lim {\varepsilon \rightarrow 0} e(\varepsilon)=0, \lim {R \rightarrow \infty} E(R)=0$. It follows that
$$
\left|\frac{\pi}{2}-\int_{\varepsilon}^{R} \frac{\sin x}{x} d x\right| \leq\left|\frac{\pi}{2}-\int_{0}^{R} \frac{1}{1+t^{2}} d t\right|+|e(\varepsilon)|+|E(R)|
$$
Since $\lim {R \rightarrow \infty} \int{0}^{R} \frac{1}{1+t^{2}} d t=\frac{\pi}{2}$, It follows that if $\varepsilon$ is sufficiently small and positive and if $R$ is sufficiently large and positive, then $\left|\frac{\pi}{2}-\int_{\varepsilon}^{R} \frac{\sin x}{x} d x\right|<\varepsilon$ Thus $\frac{\pi}{2}=\int_{0}^{\infty} \frac{\sin (x)}{x} d x$ as claimed.
A much shorter way to verify this identity is in the exercises but it depends on a theorem which has not been discussed yet and to use it, you need to know the existence of the Dirichlet integral which is obtained here as part of the argument.
For $I$ an interval let
$$
\mathcal{X}{I}(t) \equiv\left{\begin{array}{l} 1 \text { if } t \in I \ 0 \text { if } t \notin I \end{array}\right. $$ Lemma 10.1.2 Suppose $f$ is Riemann integrable on $[a, b]$. Then for each $\varepsilon>0$, there is a step function s which satisfies $|s(x)| \leq|f(x)|$ and $$ \int{a}^{b}|f(x)-s(x)| d x<\varepsilon
$$
Also there exists a continuous function $h$ which is 0 at a and b such that $|h| \leq|f|$ and
$$
\int_{a}^{b}|f(x)-h(x)|^{2} d x<\varepsilon
$$
Proof: First suppose $f(x) \geq 0$. Then by Theorem $9.3 .10$, there is a lower sum $L(f, P)$ such that
$$
\left|\int_{a}^{b} f d t-L(f, P)\right| \leq(U(f, P)-L(f, P))<\varepsilon
$$
Let $s$ correspond to this lower sum. That is, if $L(f, P)=\sum_{k=1}^{m} m_{k}\left(x_{k}-x_{k-1}\right)$, you let $s(x) \equiv \sum_{k=1}^{m} m_{k} \mathcal{X}{I{k}}(x)$ where $I_{1}=\left[x_{0}, x_{1}\right], I_{k}=\left(x_{k-1}, x_{k}\right]$ for $k \geq 2$. Then
$$
\int_{a}^{b}|f(x)-s(x)| d x=\int_{a}^{b}(f(x)-s(x)) d x=\int_{a}^{b} f d t-L(f, P)<\varepsilon
$$
Definition $10.0 .1$ The term $L^{1}(a, b)$ where $a \geq-\infty$ and $b \leq \infty$ will refer to functions which are Riemann integrable on finite intervals properly contained in $(a, b)$ and $\lim {r \rightarrow a+, R \rightarrow b-} \int{r}^{R}|f(t)| d t$ exists and the value of this limit will be denoted by $\int_{a}^{b}|f(t)| d t$. The symbol means that whenever $\varepsilon>0$, there exists $\tilde{r}>a$ and $\tilde{R}\tilde{R}$, you have the number denoted as $\int_{a}^{b}|f(t)| d t$ satisfying
$$
\left|\int_{a}^{b}\right| f(t)\left|d t-\int_{r}^{R}\right| f(t)|d t|<\varepsilon
$$
(When you have a decent integral, this whole presentation is a lot easier.) The claims of the following example are easy and are left for you to verify.
Example 10.0.2 If $f(x)=x^{-p}$, for $p \in(0,1)$, then $f \in L^{1}(0,1)$. If $f(x)=e^{-x}$, then $f \in L^{1}(-a, \infty)$ for any $a<\infty$. If $f(x)=1 / x^{2}$, then $f \in L^{1}(1, \infty)$. If $f(x)=e^{-t} t^{-p}$ where $p>0$, then $f \in L^{1}(0, \infty)$.
Lemma 10.0.3 Suppose $f \in L^{1}(a, b)$. Then for $g(t)$ any bounded continuous function, say $|g(t)| \leq M$
$$
\lim {r \rightarrow a+, R \rightarrow b-} \int{r}^{R} f(t) g(t) d t \text { exists }
$$
and this will be denoted as $\int_{a}^{b} f(t) g(t) d t$. It is called the improper integral. Also, it can be computed by letting $r_{n}$ be a decreasing sequence converging to a and $R_{n}$ an increasing sequence converging to $b$ and then
$$
\int_{a}^{b} f(t) g(t) d t=\lim {n \rightarrow \infty} \int{r_{n}}^{R_{n}} f(t) g(t) d t
$$
Proof: There is no issue about the existence of $\int_{r}^{R} f(t) g(t) d t$ thanks to Theorem $9.3 .12$. Let $\tilde{R}, \tilde{r}$ be as described in the definition corresponding to $\frac{\varepsilon}{5 M}$ for $|f(t)|$ in place of $f(t)$. Then let $r, \hat{r}<\tilde{r}, R, \hat{R}>\tilde{R}$. Then $\left|\int_{r}^{R} f(t) g(t) d t-\int_{\hat{r}}^{\hat{R}} f(t) g(t) d t\right| \leq \int_{\min (r, \hat{r})}^{\max (r, \hat{r})}|f(t) g(t)| d t+\int_{\min (R, \hat{R})}^{\max (R, \hat{R})}|f(t) g(t)| d t$ 219 CHAPTER 10. IMPROPER INTEGRALS
$$
\begin{aligned}
&\leq M \int_{\min (r, \hat{r})}^{\bar{r}}|f(t)| d t+M \int_{\bar{R}}^{\max (R, \hat{R})}|f(t)| d t \
&\leq M\left(\int_{r}^{\bar{r}}|f(t)| d t+\int_{\bar{R}}^{R}|f(t)| d t+\int_{\hat{r}}^{\bar{r}}|f(t)| d t+\int_{\bar{R}}^{\hat{R}}|f(t)| d t\right)<4 M \frac{\varepsilon}{5 M}=\frac{4 \varepsilon}{5} \end{aligned} $$ Thus, letting $\varepsilon_{n}$ decrease to 0 , there exists $\tilde{r}_{n}, \tilde{R}_{n}$ with $\tilde{r}_{n}$ decreasing to $a$ and $\tilde{R}_{n}$ increasing to $b$ such that the diameters of the set $$ S_{n} \equiv \overline{\left\{\int_{r}^{R} f(t) g(t) d t: aR>\tilde{R}{n}\right}} $$ is no more than $\frac{4}{5} \varepsilon{n}$. By completeness, there is $I \in \cap_{n} S_{n}$. Then if $r<\tilde{r}{n}, R>\tilde{R}{n}$, $$ \left|I-\int_{r}^{R} f(t) g(t) d t\right| \leq \frac{4}{5} \varepsilon_{n}<\varepsilon_{n} $$ and so the improper integral exists and equals $I$. Let $\varepsilon>0$ be given and let $\varepsilon_{n}
<\varepsilon .$ Let $r_{k}$ decrease to $a$ and $R_{k}$ increase to
$b$, then for $k$ large enough, $r_{k}<\tilde{r}{n}$ and $\bar{R} {n}<R_{k}$ so $\int_{r_{k}}^{R_{k}} f(t) g(t) d t \in S_{n}$ and hence
$\left|I-\int_{r_{k}}^{R_{k}} f(t) g(t) d t\right| \leq \frac{4 \varepsilon}
{5}<\varepsilon$ which shows that
$$
\begin{aligned}
&=e(\varepsilon)+E(R)+\int_{0}^{R} \frac{1}{1+t^{2}} d t-\int_{0}^{R} e^{-R t} \frac{(\cos R)+t(\sin R)}{1+t^{2}} d t \
&=e(\varepsilon)+E(R)+\int_{0}^{R} \frac{1}{1+t^{2}} d t-\int_{0}^{R} e^{-R t} \frac{1}{\sqrt{1+t^{2}}} \cos \left(R-\phi_{t}\right) d t
\end{aligned}
$$
where $\phi_{t}$ is a phase shift. This is because
$$
\left(\frac{1}{\sqrt{1+t^{2}}}, \frac{t}{\sqrt{1+t^{2}}}\right)=\left(\cos \phi_{t}, \sin \phi_{t}\right) \text { some } \phi_{t}
$$
so the formula for $\cos (R-\phi)$ is used. This last integral satisfies
$$
\left|\int_{0}^{R} e^{-R t} \frac{1}{\sqrt{1+t^{2}}} \cos \left(R-\phi_{t}\right) d t\right| \leq \int_{0}^{R} e^{-R t} d t \leq \frac{1}{R}
$$
Therefore, include it in $E(R)$ and it follows that
$$
\int_{\varepsilon}^{R} \frac{\sin x}{x} d x=e(\varepsilon)+E(R)+\int_{0}^{R} \frac{1}{1+t^{2}} d t
$$
where $\lim {\varepsilon \rightarrow 0} e(\varepsilon)=0, \lim {R \rightarrow \infty} E(R)=0$. It follows that
$$
\left|\frac{\pi}{2}-\int_{\varepsilon}^{R} \frac{\sin x}{x} d x\right| \leq\left|\frac{\pi}{2}-\int_{0}^{R} \frac{1}{1+t^{2}} d t\right|+|e(\varepsilon)|+|E(R)|
$$
Since $\lim {R \rightarrow \infty} \int{0}^{R} \frac{1}{1+t^{2}} d t=\frac{\pi}{2}$, It follows that if $\varepsilon$ is sufficiently small and positive and if $R$ is sufficiently large and positive, then $\left|\frac{\pi}{2}-\int_{\varepsilon}^{R} \frac{\sin x}{x} d x\right|<\varepsilon$ Thus $\frac{\pi}{2}=\int_{0}^{\infty} \frac{\sin (x)}{x} d x$ as claimed.
A much shorter way to verify this identity is in the exercises but it depends on a theorem which has not been discussed yet and to use it, you need to know the existence of the Dirichlet integral which is obtained here as part of the argument.
For $I$ an interval let
$$
\mathcal{X}{I}(t) \equiv\left{\begin{array}{l} 1 \text { if } t \in I \ 0 \text { if } t \notin I \end{array}\right. $$ Lemma 10.1.2 Suppose $f$ is Riemann integrable on $[a, b]$. Then for each $\varepsilon>0$, there is a step function s which satisfies $|s(x)| \leq|f(x)|$ and $$ \int{a}^{b}|f(x)-s(x)| d x<\varepsilon
$$
Also there exists a continuous function $h$ which is 0 at a and b such that $|h| \leq|f|$ and
$$
\int_{a}^{b}|f(x)-h(x)|^{2} d x<\varepsilon
$$
Proof: First suppose $f(x) \geq 0$. Then by Theorem $9.3 .10$, there is a lower sum $L(f, P)$ such that
$$
\left|\int_{a}^{b} f d t-L(f, P)\right| \leq(U(f, P)-L(f, P))<\varepsilon
$$
Let $s$ correspond to this lower sum. That is, if $L(f, P)=\sum_{k=1}^{m} m_{k}\left(x_{k}-x_{k-1}\right)$, you let $s(x) \equiv \sum_{k=1}^{m} m_{k} \mathcal{X}{I{k}}(x)$ where $I_{1}=\left[x_{0}, x_{1}\right], I_{k}=\left(x_{k-1}, x_{k}\right]$ for $k \geq 2$. Then
$$
\int_{a}^{b}|f(x)-s(x)| d x=\int_{a}^{b}(f(x)-s(x)) d x=\int_{a}^{b} f d t-L(f, P)<\varepsilon
$$
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