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example of improper integral
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(Example)
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The integrand of
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(1) |
is undefined both at the lower and the upper limit. However, the value of the improper integral exists and may be found via the more general integral
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Denote the integrand of (2) by $f(x,\,y)$ . For any fixed real value $y$ , $$f(x,\,y) \in O(1) \mbox{\; as \;} x \to 0, \quad f(x,\,y) \in O(\frac{1}{\sqrt{1\!-\!x^2}}) \mbox{\; as \;} x \to 1,$$ where the Landau big O notation has been used. Accordingly, the integral (2) converges for every $y$ .
The inequality $$\left|\frac{\partial f(x,\,y)}{\partial y}\right| \;=\; \frac{1}{(1\!+\!x^2y^2)\sqrt{1\!-\!x^2}} \;\leqq\; \frac{1}{\sqrt{1\!-\!x^2}}$$ and the convergence of the integral $$\int_0^1\!\frac{dx}{\sqrt{1\!-\!x^2}} \;=\; \frac{\pi}{2}$$ imply that the integral
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(3) |
converges uniformly on the whole $y$ -axis and equals $I'(y)$ . For expressing this derivative in a closed form, one may utilise the changes of variable $$x \;:=\; \cos\varphi, \quad \tan\varphi \;:=\; t$$ which yield
Hence, $$I(y) \;=\; \frac{\pi}{2}\int_0^y\frac{dy}{\sqrt{1\!+\!y^2}} \;=\; \sijoitus{0}{\quad y}\!\ln(y+\sqrt{1\!+\!y^2})$$ and the integral (1) equals $I \;=\; I(1) \;=\; \frac{\pi}{2}\ln(1\!+\!\sqrt{2})$ , i.e.
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(4) |
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"example of improper integral" is owned by pahio.
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Cross-references: derivative, imply, inequality, converges, integral, real, fixed, general integral, improper integral, upper limit, integrand
There are 2 references to this entry.
This is version 2 of example of improper integral, born on 2009-02-11, modified 2009-02-15.
Object id is 11617, canonical name is ExampleOfImproperIntegral.
Accessed 565 times total.
Classification:
| AMS MSC: | 40A10 (Sequences, series, summability :: Convergence and divergence of infinite limiting processes :: Convergence and divergence of integrals) |
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Pending Errata and Addenda
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