1.4 Integrals of the Second Kind
Similar convergence theorems as those of the integrals of the first kind can be formulated for integrals of the second kind.
The limit theorems can be formulated as \[\lim _{x\rightarrow b^-}\frac {f(x)}{g(x)}\] or \[\lim _{x\rightarrow a^+}\frac {f(x)}{g(x)}\] and so convergence of \[\int _a^b f(x)\, dx\] and \[\int _a^bg(x)\, dx\] will be the same.
For integrals of the second kind the basic standard reference integrals are \[\int _a^b\frac {1}{(b-x)^p}\, dx \hspace {0.3cm}, \hspace {0.6cm} \int _a^b\frac {1}{(x - a)^p}\, dx.\] Now \begin {align*} \int _a^b\frac {1}{(b - x)^p}\, dx & = -\frac {[(b - x)^{1 - p}]}{1-p}\Bigg |_a^b\\ & = \frac {(b -a)^{1-p}}{1-p} - \lim _{x \rightarrow b^-} \frac {(b-x)^{1-p}}{1 - p}\\ & = \frac {(b-a)^{1-p}}{1 - p}\, , \hspace {0.5cm}\text {when}\,\, 1 - p > 0\,\, \text {or}\,\, p < 1. \end {align*}
And so the integral will diverge when \(p\geq 1\).
If \(p < 0\), the integrals are proper with no singularities of the integrals.
- (a).
- The integral
\[\int _0^1\frac {1}{(1 - x^3)^{\frac {1}{3}}}\, dx\]
is convergent.
Working. \begin {align*} f(x) & = \frac {1}{(1 - x^3)^{\frac {1}{3}}}\\ & = \frac {1}{\left [(1 - x)\, (1 + x + x^2)\right ]^{\frac {1}{3}}}\\ & = \frac {1}{(1 - x)^{\frac {1}{3}}\, (1 + x + x^2)^{\frac {1}{3}}}. \end {align*}
So let \[g(x) = \frac {1}{(1 - x)^{\frac {1}{3}}}.\] Then \[\lim _{x\rightarrow 1^-} \frac {f(x)}{g(x)} = \lim _{x\rightarrow 1^-} \frac {1}{(1 + x + x^2)^{\frac {1}{3}}} = \frac {1}{3^{\frac {1}{3}}}.\] Since \[\int _a^bg(x)\, dx\] converges \((p = \frac {1}{3})\), so also is \[\int _a^bf(x)\, dx.\] □
- (b).
- Determine convergence or divergence by comparing with
\[\int _0^1\frac {1}{x^p}\, dx \, , \hspace {0.6cm} \int _0^1\frac {1}{(b - x)^2}\, dx\]
- (i).
- \(\displaystyle {\int _0^1 \frac {(1 + 2x)\, \sqrt {1 + x^2}}{1 - x^2}}\, dx\)
- (ii).
- \(\displaystyle {\int _0^1 \frac {1}{\sqrt {x}\, (x + 2x^2)^{\frac {1}{3}}}}\, dx\)
- (iii).
- \(\displaystyle {\int _1^2\frac {4(8 - x^3)}{2(x - x^2)^2}}\, dx\)
- (iv).
- \(\displaystyle {\int _0^1\left (\frac {1}{x} + x\right )^{\frac {1}{2}\,\frac {x^{\frac {1}{3}}}{x - \frac {1}{2}x^2}}}\, dx\)
Working.
- (i).
- For (i) \[\frac {(1 + 2x)\, \sqrt {1 + x^2}}{1 - x^2} = \frac {(1 + 2x) \, \sqrt {1 + x^2}}{(1 - x)(1 + x)} = f(x).\] Set \[g(x) = \frac {1}{1 - x}\] (diverges)
- (ii).
- \[f(x) = \frac {1}{\sqrt {x}\, (x + 2x^2)} = \frac {1}{x^{\frac {3}{2}}\, (1 + 2x)}.\] Set \[g(x) = \frac {1}{x^{\frac {3}{2}}}\] (diverges, since \(p = \frac {3}{2} >1\))
- (iii).
- \[\frac {4(8-x^3)}{2(x - x^2)^2} = \frac {2(8 - x^3)}{x^2(1 - x)^2}.\] Set \[g(x) = \frac {1}{(1-x)^2}\] since \(p = 2 > 1\) diverges.
- (iv).
- \[\left (\frac {1}{x} + x\right )^{\frac {1}{2}} \, \frac {x^{\frac {1}{3}}}{x - \frac {1}{2}x^2}.\] Set \[g(x) = \frac {1}{x^{\frac {3}{2}}} = \frac {\left (\frac {1 + x^2}{x}\right )^{\frac {1}{2}} x^{\frac {1}{3}}}{x^{\frac {3}{2}} \left (1 - \frac {1}{2}x\right )},\] diverges since \(p = \frac {3}{2} > 1\).
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