1.2 Improper Integrals of The Second Kind
An integral of the form \[\int _a^b f(x)\, dx,\] with finite limits is said be of the second kind if the failure to be an ordinary proper integral arises from the behavior of \(f(x)\) as \(x\rightarrow a\) or as \(x\rightarrow b\) but not both. Thus, if \(f(x)\) is integrable on \([a,c]\) for each \(c\) such that \(a<c<b\), but is not integrable on \([a,b]\), we say the integrable is improper at \(x = b\). We also say \(f(x)\) has a singularity at \(x = b\). We then define \[\int _a^bf(x)\, dx = \lim _{c\rightarrow a^-}\int _a^cf(x)\, dx\] so the improper integral will converge or diverge according as the limit exits or not. Similarly, if there is a singularity at \(x = a\), we define \[\int _a^bf(x)\, dx = \lim _{c\rightarrow a^+}\int _c^bf(x)\, dx.\]
Example 1.2.1. \[\int _0^1\frac {1}{\sqrt {1 - x^2}}\, dx\, , \hspace {2cm} \int _0^{\frac {\pi }{3}}\frac {1}{\sqrt {\cos \theta - \frac {1}{2}}}\, dx\, , \hspace {2cm} \int ^1_{\frac {1}{2}}\frac {\log x}{(1 - x)^{3/2}}\, dx\] are improper at the upper limits and \[\int _0^1\left [\log \left (\frac {1}{x}\right )\right ]^3\, dx \, , \hspace {2cm} \int _0^1\frac {e^{-x}}{\sqrt {x}}\, dx\, , \hspace {2cm} \int _0^1\frac {\log x}{1 + x}\, dx\] and improper at the lower limits
Example 1.2.2. Investigate the improper integrals
- (a).
- \(\displaystyle {\int _0^1\frac {1}{\sqrt {x}}\, dx}\)
- (b).
- \(\displaystyle {\int _1^2\frac {1}{(x - 2)^2}\, dx}\)
- (c).
- \(\displaystyle {\int _0^3\frac {1}{x - 1}\, dx}\)
- (d).
- \(\displaystyle {\int _0^2\frac {1}{(2x - 1)^{\frac {2}{3}}}}\)
- (e).
- \(\displaystyle {\int _0^1 \ln x\, dx}\)
Working.
- (a).
- \(\displaystyle {\int _0^1 \frac {1}{\sqrt {x}}}\) \begin {align*} \int _0^1 \frac {1}{\sqrt {x}}\, dx & = \lim _{t\rightarrow 0^+}\int _t^1\frac {1}{x}\, dx\\ & = \lim _{t\rightarrow 0^+}\, 2\sqrt {x}\big |_t^1\\ & = \lim _{t\rightarrow 0^+}\left [2 - 2\sqrt {2}\right ]\\ & = 2. \end {align*}
converges.
- (b).
- \(\displaystyle {\int _1^2\frac {1}{(x - 2)^2}\, dx}\) \begin {align*} \int _1^2\frac {1}{(x - 2)^2}\, dx & = \lim _{t\rightarrow 2^-}\int _1^t\frac {1}{(x-2)^2}\, dx\\ & = \lim _{t\rightarrow 2^-} \left [-\frac {1}{x - 2}\right ]_1^t\\ & = \lim _{t\rightarrow 2^-} \left [-\frac {1}{t - 2} - 1\right ]\\ & = +\infty , \end {align*}
so integral diverges.
- (c).
- \(\displaystyle {\int _0^3\frac {1}{x - 1}\, dx}\) \begin {align*} \int _0^3\frac {1}{x - 1}\, dx & = \int _0^1\frac {1}{x - 1}\, dx + \int _1^3\frac {1}{x - 1}\, dx\\ & = \lim _{t\rightarrow 1^-}\int _0^t \frac {1}{x - 1}\, dx + \lim _{t\rightarrow 1^+}\int _t^3\frac {1}{x - 1}\, dx\\ & = -\infty + \infty , \end {align*}
so integral diverges.
observe that if you ignored the singularity at \(x = 1\), we would have \[\int _0^3\frac {1}{x - 1}\, dx = \ln |x - 1|\big |_0^3 = \ln 2 - \ln 1 = \ln 2,\] which would be wrong. (Because the integrand is not continuous).
Exercise 1.2.3. Investigate the convergence of the following:
- (a).
- \(\displaystyle {\int ^{\infty }_1\frac {1}{(3x + 1)^2}\, dx}\)
- (b).
- \(\displaystyle {\int _2^{\infty } \frac {1}{(x + 3)^{\frac {3}{2}}}\, dx}\)
- (c).
- \(\displaystyle {\int ^{-1}_{-\infty } e^{-2t}\, dt}\)
- (d).
- \(\displaystyle {\int _1^{\infty } \frac {\ln x}{x}}\ dx\)
- (e).
- \(\displaystyle {\int _0^{\infty } \frac {1}{x\sqrt {x}}\, dx}\)
- (f).
- \(\displaystyle {\int _2^{\infty } \frac {1}{2\sqrt {x}}\, dx}\)
- (g).
- \(\displaystyle {\int _{-\infty }^{\infty }\frac {x}{x^2 + 4}\, dx}\)
- (h).
- \(\displaystyle {\int _{-\infty }^{\infty } |x|\, e^{-x^2}\, dx}\)
- (i).
- \(\displaystyle {\int _0^{\infty } \frac {1}{x + x^2}\, dx}\)
- (j).
- \(\displaystyle {\int _2^{\infty } \frac {1}{x\, \left [\ln x\right ]^2}\, dx}\)
- (k).
- \(\displaystyle {\int ^{\infty }_{-\infty } \frac {x}{(x^2 + 4)^{\frac {3}{2}}}\, dx}\)
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