689. Find the following improper integrals:
1) $\int_{0}^{+\infty} e^{-x} d x$
2) $\int_{-\infty}^{+\infty} \frac{d x}{x^{2}+1}$
3) $\int_{0}^{1} \frac{d x}{x}$
4) $\int_{-1}^{2} \frac{d x}{\sqrt[3]{(x-1)^{2}}}$

Explain the solutions geometrically.
Solution. 1) Using equality (1), we have

$$
\int_{0}^{+\infty} e^{-x} d x=\lim _{\beta \rightarrow+\infty} \int_{0}^{\beta} e^{-x} d x=\left.\lim \left(-e^{-x}\right)\right|_{0} ^{\beta}=\lim \left(e^{0}-e^{-\beta}\right)=1
$$

Therefore, the given improper integral converges.

Geometrically, in a rectangular coordinate system, any definite integral $\int_{a}^{b} f(x) d x$ gives the algebraic sum of the areas bounded by the curve $y=f(x)$, two vertical lines $x=a, x=b$, and the x-axis. Therefore, by plotting the curve $y=e^{-x}$ and its ordinates at points $x=0$ and $x=\beta$ (Fig. 128), we obtain the curvilinear trapezoid $O A B \beta$, the area of which is

$$
S(\beta)=\int_{0}^{\beta} e^{-x} d x=1-e^{-\beta}
$$

As $\beta \rightarrow+\infty$, we obtain a trapezoid with an infinite base, which has a finite area $S(+\infty)=\lim _{\beta \rightarrow+\infty} S(\beta)=1$.

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-226.jpg?height=257&width=471&top_left_y=359&top_left_x=128)

Fig. 128

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-226.jpg?height=227&width=484&top_left_y=389&top_left_x=656)

Fig. 129

2) Using definition (3), we get

$$
\begin{gathered}
\int_{-\infty}^{+\infty} \frac{d x}{x^{2}+1}=\lim _{a \rightarrow-\infty} \int_{a}^{0} \frac{d x}{x^{2}+1}+\lim _{\beta \rightarrow+\infty} \int_{0}^{\beta} \frac{d x}{x^{2}+1}=\left.\lim \operatorname{arctg} x\right|_{a} ^{0}+ \\
+\left.\lim \operatorname{arctg} x\right|_{0} ^{\beta}=-\operatorname{arctg}(-\infty)+\operatorname{arctg}(+\infty)= \\
=-\left(-\frac{\pi}{2}\right)+\frac{\pi}{2}=\pi
\end{gathered}
$$

Geometrically (Fig. 129), the integral of the function $f(x)=\frac{1}{x^{2}+1}$ from $\alpha$ to $\beta$ represents the area of the curvilinear trapezoid $\alpha A B \beta$, and the given convergent improper integral represents the area of the infinite curvilinear trapezoid that extends infinitely to the left and right and has a finite value of $\pi$.

3) Here, the integrand function $\frac{1}{x}$ has an infinite discontinuity at $x=0$. According to definition (4),

$$
\int_{0}^{1} \frac{d x}{x}=\lim _{\varepsilon \rightarrow+0} \int_{\varepsilon}^{1} \frac{d x}{x}=\left.\lim \ln x\right|_{\varepsilon} ^{1}=\lim (\ln 1-\ln \varepsilon)=-\ln 0=+\infty
$$

i.e., this improper integral diverges.

Geometrically (Fig. 130), the obtained result indicates that the area of the curvilinear trapezoid $\varepsilon A B b$

$$
S(\varepsilon)=\int_{\varepsilon}^{1} \frac{d x}{x}=-\ln \varepsilon
$$

unboundedly increases as $\varepsilon \rightarrow+0$.

4) Here, the integrand function has an infinite discontinuity at $x=1$, which lies within the integration interval $[-1, 2]$. Therefore, according to definition (4),

$$
\int_{-1}^{2} \frac{d x}{\sqrt[3]{(x-1)^{2}}}=\lim _{\varepsilon_{1} \rightarrow+0} \int_{-1}^{1-\varepsilon_{1}} \frac{d x}{\sqrt[3]{(x-1)^{2}}}+\lim _{\varepsilon_{2} \rightarrow+0} \int_{1+\varepsilon_{2}}^{2} \frac{d x}{\sqrt[3]{(x-1)^{2}}}=
$$

$$
\begin{gathered}
=\left.\lim 3 \sqrt[3]{x-1}\right|_{-1} ^{1-\varepsilon_{1}}+\left.\lim 3 \sqrt[3]{x-1}\right|_{1+\varepsilon_{2}} ^{2}=3 \lim \left(\sqrt[3]{-\varepsilon_{1}}-\sqrt[3]{-2}\right)+ \\
+3 \lim \left(\sqrt[3]{1}-\sqrt[3]{\varepsilon_{2}}\right)=3(\sqrt[3]{2}+1)
\end{gathered}
$$

For the graph of the integrand function $y=\frac{1}{\sqrt[3]{(x-1)^{2}}}$ (Fig. 131), the line $x=1$ is a vertical asymptote. The integrals of this function from -1 to $1-\varepsilon_{1}$ and from $1+\varepsilon_{2}$ to 2 represent the areas of the curvilinear trapezoids $a A P \varepsilon$ and $\eta Q B b$. As $\varepsilon_{1} \rightarrow+0$ and $\varepsilon_{2} \rightarrow+0$, these trapezoids extend infinitely upwards and yet have finite areas, the sum of which equals the value of the given convergent improper integral.

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-227.jpg?height=454&width=203&top_left_y=878&top_left_x=284)

Fig. 130

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-227.jpg?height=378&width=403&top_left_y=946&top_left_x=649)

Fig. 131