1.1 Improper Integrals of the First Kind
An integral of the form
\[\int _a^{\infty } f(x)\, dx\]
in which \(f(x)\) is defined for \(x\geq a\) and integrable over every finite interval \([a,b]\) is said to be of the first kind.
We then have
\[\int _a^{\infty }f(x)\, dx = \lim _{b\rightarrow \infty }\int _a^bf(x)\, dx.\]
If the limit exits we say that the improper integral is convergent, otherwise it is said to be
divergent.
If the function \(f\) is continuous on the whole real line, we define \[\int _{-\infty }^{\infty }f(x)\, dx = \int _{-\infty }^cf(x)\, dx + \int _c^{\infty } f(x)\, dx\] for any convenient choice of \(c\), provided the two integrals on the right hand side both converge.
Note that the choice of \(c\) above does not matter. For, if \(c<d\) \begin {align*} \int _{-\infty }^cf(x)\, dx + \int _c^{\infty }f(x)\, dx & = \int _{-\infty }^cf(x)\, dx + \int _c^df(x)\, dx + \int _d^{\infty }f(x)\, dx\\ & = \int ^d_{-\infty }f(x)\, dx + \int _d^{\infty }f(x)\, dx. \end {align*}
Also we immediately observe that \[\int _{-\infty }^{\infty }f(x)\, dx \neq \lim _{t\rightarrow \infty } \int ^t_{-t} f(x)\, dx.\]
Example 1.1.1. Show that \[\int _{-\infty }^{\infty } \frac {1 + x}{1 + x^2}\, dx\] diverges, but that \[\lim _{t\rightarrow \infty }\int _{-t}^{t} \frac {1 + x}{1 + x^2}\, dx = \pi .\]
Working. First consider the indefinite integral \begin {align*} \int \frac {1 + x}{1 + x^2}\, dx & = \int \frac {1}{1 + x^2}\, dx + \int \frac {x}{1 + x^2}\, dx\\ & = \tan ^{-1}x + \frac {1}{2}\ln (1 + x^2) + c. \end {align*}
Now \begin {align*} \int ^{\infty }_{-\infty } \frac {1 + x}{1 + x^2}\,\, dx & = \int ^{\infty }_{c} \frac {1 + x}{1 + x^2}\, dx + \int ^{c}_{-\infty } \frac {1 + x}{1 + x^2}\, dx\\\\ & = \lim _{t\rightarrow \infty }\int ^{t}_{c} \frac {1 + x}{1 + x^2}\, dx + \lim _{t\rightarrow \infty }\int ^{c}_{-t} \frac {1 + x}{1 + x^2}\, dx\\ & = \infty - \infty , \end {align*}
so the integral diverges.
However, \begin {align*} \lim _{t\rightarrow \infty } \int _{-t}^t \frac {1 + x}{1 + x^2}\, dx & = \lim _{t\rightarrow \infty }\left [\tan ^{-1}x + \frac {1}{2} \ln (1 + x^2) \right ]_{-t}^t\\ & = \lim _{t\rightarrow \infty }\left [\tan ^{-1}(t) - \tan ^{-1}(-t)\right ]\\ & = \frac {\pi }{2} - \left (\frac {-\pi }{2}\right )\\ & = \pi . \end {align*} □
Example 1.1.2. Investigate the improper integrals
- (a).
- \(\displaystyle {\int _1^{\infty } \frac {1}{x^2}\, dx}\)
- (b).
- \(\displaystyle {\int _1^{\infty }\frac {1}{x}\, dx}\)
- (c).
- \(\displaystyle {\int _{-\infty }^0\frac {1}{\sqrt {1 - x}}\, dx}\)
Working.
- (a).
- \[\int ^{\infty }_1\frac {1}{x^2}\, dx = \lim _{t\rightarrow \infty } \int _1^t\frac {1}{x^2}\, dx = \lim _{t\rightarrow \infty }\left [-\frac {1}{x}\right ]_1^t = \lim _{t\rightarrow \infty }\left [-\frac {1}{t} + 1\right ] = 1.\] So the integral converges.
- (b).
- \[\int _1^{\infty }\frac {1}{x}\, dx = \lim _{t\rightarrow \infty } \int _1^t\frac {1}{x}\, dx = \lim _{t\rightarrow \infty } \left [\ln |x|\right ]_1^t = \lim _{t\rightarrow \infty }[\ln t - \ln 1] = \infty ,\] so the integral diverges.
- (c).
- \[\int _{-\infty }^0\frac {1}{\sqrt {1 - x}}\, dx = \lim _{t\rightarrow -\infty }\int _t^0\frac {1}{\sqrt {1 - x}}\, dx = \lim _{t\rightarrow -\infty }\left [-2\sqrt {1 - x}\right ]_t^0 = \infty ,\] diverges.
Example 1.1.3 (\(p\)-Integrals). For what values of \(p\) is the integral \[\int _1^{\infty }\frac {1}{x^p}\, dx\] convergent?
Working. \[\int _1^{\infty }\frac {1}{x^p}\, dx = \lim _{t\rightarrow \infty }\int _1^t\frac {1}{x^p} = \lim _{t\rightarrow \infty }\left [\frac {1}{1 - p}(t^{1 - p} - 1)\right ].\] Now if \(p = 1\), we have seen in example 1.1.2 that the integral diverges.
So if \(p>1\), then \(1 - p < 0\) so \(t^{1-p}\longrightarrow 0\) as \(t\longrightarrow \infty \), hence \[\int _1^{\infty }\frac {1}{x^p}\, dx = \frac {1}{p - 1}\hspace {0.3cm} \text {when}\hspace {0.3cm} p>1.\] If \(p < 1\), then \(1 - p > 0\) and so \(t^{1 - p}\longrightarrow \infty \) as \(t\longrightarrow \infty \) and so the improper integral diverges. □
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