1334. Find the improper integrals:
1) $\int_{0}^{+\infty} e^{-x} d x$
2) $\int_{-\infty}^{+\infty} \frac{d x}{1+x^{2}}$
3) $\int_{1}^{+\infty} \frac{d x}{x}$
4) $\int_{1}^{+\infty} \frac{d x}{x^{2}}$
Solution. 1) The integrand is continuous on the interval $[0,+\infty)$, hence by definition $\int_{0}^{+\infty} e^{-x} d x=\lim _{b \rightarrow+\infty} \int_{0}^{b} e^{-x} d x=\left.\lim _{b \rightarrow+\infty}\left(-e^{-x}\right)\right|_{0} ^{b}=\lim _{b \rightarrow+\infty}\left(-e^{-b}+1\right)=1$.

It is known that the area of a curvilinear trapezoid bounded by the graph of the function $y=f(x)$, the lines $x=a, y=b$, and the segment of the $O x$ axis is expressed by the integral $\int_{a}^{b} f(x) d x$. This is the geometric meaning of the definite integral (see § 4). If the integral is improper and convergent, then its geometric meaning is the same as that of a proper definite integral. Thus, by computing this improper convergent integral, we obtained the value of the area of the plane figure bounded by the graph of the function $f(x)=e^{-x}$, the line $x=0$, and the $O x$ axis (Fig. $93, a)$.

![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-262.jpg?height=314&width=884&top_left_y=551&top_left_x=158)

Fig. 93

2) By definition we have

$$
\int_{-\infty}^{+\infty} \frac{d x}{1+x^{2}}=\lim _{a \rightarrow-\infty} \int_{a}^{0} \frac{d x}{1+x^{2}}+\lim _{b \rightarrow+\infty} \int_{0}^{b} \frac{d x}{1+x^{2}}=\left.\lim _{a \rightarrow-k} \operatorname{arctg} x\right|_{a} ^{0}+
$$

$+\left.\lim _{b \rightarrow+\infty} \operatorname{arctg} x\right|_{0} ^{b}=-\operatorname{arctg}(-\infty)+\operatorname{arctg}(+\infty)=-\left(-\frac{\pi}{2}\right)+$

$$
+\frac{\pi}{2}=\pi
$$

Geometrically, this improper convergent integral represents the area of the curvilinear trapezoid bounded by the graph of the integrand and the $O x$ axis, which is a horizontal asymptote (Fig. 93, b).

3) The integrand is continuous on the interval $[1,+\infty)$, hence,

$$
\int_{1}^{+\infty} \frac{d x}{x}=\lim _{b \rightarrow+\infty} \int_{i}^{b} \frac{d x}{x}=\left.\lim _{b \rightarrow+\infty} \ln x\right|_{1} ^{b}=\lim _{b \rightarrow+\infty}(\ln b-\ln 1)=+\infty
$$

This integral diverges.

$$
\text { 4) } \int_{1}^{+\infty} \frac{d x}{x^{2}}=\lim _{b \rightarrow+\infty} \int_{1}^{b} \frac{d x}{x^{2}}==\left.\lim _{b \rightarrow+\infty}\left(-\frac{1}{x}\right)\right|_{1} ^{b}=\lim _{b \rightarrow+\infty}\left(-\frac{1}{b}+1\right)=1
$$

The integral converges and represents the area of the curvilinear trapezoid bounded by the lines $x=1, y=0$ and the graph of the function $y=\frac{1}{x^{2}}$.