Circular trap with integration by parts - Revision history

Vipul: /* Product of inverse trigonometric or logarithmic function and rational function with integral that is inverse trigonometric or logarithmic */

2012-04-14T03:14:50Z

Product of inverse trigonometric or logarithmic function and rational function with integral that is inverse trigonometric or logarithmic

← Older revision		Revision as of 03:14, 14 April 2012
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	ILATE suggests that we take <math>\arctan</math> (the inverse trigonometric function) as the part to differentiate and <math>1/x</math> (the algebraic function) as the part to integrate. We get:		ILATE suggests that we take <math>\arctan</math> (the inverse trigonometric function) as the part to differentiate and <math>1/x</math> (the algebraic function) as the part to integrate. We get:

	<math>\int \frac{\arctan x}{x} \, dx = (\arctan x)(\ln x) - \int \frac{\ln x \, dx}{x^2 + 1}</math>		<math>\int \frac{\arctan x}{x} \, dx = (\arctan x)(\ln \|x\|) - \int \frac{\ln \|x\| \, dx}{x^2 + 1}</math>

	Now, ILATE suggests that for the integral on the right side, we take <math>\ln x</math> (the logarithmic function) as the part to differentiate) and <math>1/(x^2 + 1)</math> as the part to integrate. ''However'', we immediately see that making this choice would fall into the circular trap.		Now, ILATE suggests that for the integral on the right side, we take <math>\ln \|x\|</math> (the logarithmic function) as the part to differentiate) and <math>1/(x^2 + 1)</math> as the part to integrate. ''However'', we immediately see that making this choice would fall into the circular trap.

	In fact, there is no other reasonable choice to make, so this integral cannot be done using the standard function integrations. Note, however, that we ''did'' manage to establish an equivalence of integration problems:		In fact, there is no other reasonable choice to make, so this integral cannot be done using the standard function integrations. Note, however, that we ''did'' manage to establish an equivalence of integration problems:

	<math>\int \frac{\arctan x}{x} \, dx \leftrightarrow \int \frac{\ln x}{x^2 + 1} \, dx</math>		<math>\int \frac{\arctan x}{x} \, dx \leftrightarrow \int \frac{\ln \|x\|}{x^2 + 1} \, dx</math>

Vipul: /* Explanation using function notation */

2012-04-10T19:52:02Z

Explanation using function notation

← Older revision		Revision as of 19:52, 10 April 2012
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	===Explanation using function notation===		===Explanation using function notation===

			<section begin="basic"/>
	Consider an integration of the form:		Consider an integration of the form:

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	In other words, ''we ended up with the original expression, and obtained no new information in the process''. Another way of saying this is that we went in circles, i.e., went back along the path we came from, hence did not make any progress.		In other words, ''we ended up with the original expression, and obtained no new information in the process''. Another way of saying this is that we went in circles, i.e., went back along the path we came from, hence did not make any progress.
			<section end="basic"/>

	===Explanation using dependent-independent variable notation===		===Explanation using dependent-independent variable notation===

Vipul: /* Product of inverse trigonometric or logarithmic function and rational function with integral that is inverse trigonometric or logarithmic */

2012-04-10T19:35:36Z

Product of inverse trigonometric or logarithmic function and rational function with integral that is inverse trigonometric or logarithmic

← Older revision		Revision as of 19:35, 10 April 2012
Line 79:		Line 79:
	In fact, there is no other reasonable choice to make, so this integral cannot be done using the standard function integrations. Note, however, that we ''did'' manage to establish an equivalence of integration problems:		In fact, there is no other reasonable choice to make, so this integral cannot be done using the standard function integrations. Note, however, that we ''did'' manage to establish an equivalence of integration problems:

	<math>\int \frac{\arctan x}{x} \, dx = \int \frac{\ln x}{x^2 + 1} \, dx</math>		<math>\int \frac{\arctan x}{x} \, dx \leftrightarrow \int \frac{\ln x}{x^2 + 1} \, dx</math>

Vipul at 19:35, 10 April 2012

2012-04-10T19:35:00Z

← Older revision		Revision as of 19:35, 10 April 2012
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	In other words, ''we ended up with the original expression, and obtained no new information in the process''. Another way of saying this is that we went in circles, i.e., went back along the path we came from, hence did not make any progress.		In other words, ''we ended up with the original expression, and obtained no new information in the process''. Another way of saying this is that we went in circles, i.e., went back along the path we came from, hence did not make any progress.

			==Using the circular trap to illustrate the intractability of certain problems==

			===Product of polynomial and trigonometric that integrates to something involving logarithms===

			Consider the integration problem:

			<math>\! int x \tan x \, dx</math>

			If we use the ILATE precedence rule, then, we should take <math>\tan x</math> as the part to integrate and <math>x</math> as the part to differentiate. Using [[tan#Integration\|the integration of the tan function]] to be <math>-\ln\|\cos x</math>, we get:

			<math>\! \int x \tan x \, dx = -x \ln \|\cos x\| - \int (1) (-\ln \|\cos x\|) \, dx = -x\ln\|cos x\| + \int \ln\|\cos x\| \, dx</math>

			Now, we reconsider the ILATE precedence rule to integrate <math>\ln\|\cos x\|</math>. The function comes closest to being a logarithmic function, hence ILATE would suggest that we rewrite it as <math>\ln\|\cos x\|</math> times 1, picking 1 as the part to integrate. ''However'', we immediately see that making this choice would fall into the circular trap.

			Since there is no other reasonable choice to make, this problem is not tractable using integration by parts and the standard function integrations.

			Note, however, that we ''did'' manage to establish an equivalence of integration problems:

			<math>\int x \tan x \, dx \leftrightarrow \int \ln\|\cos x\| \, dx</math>

			===Product of inverse trigonometric or logarithmic function and rational function with integral that is inverse trigonometric or logarithmic===

			Consider an integral of the form:

			<math>\int \frac{\arctan x}{x} \, dx</math>

			ILATE suggests that we take <math>\arctan</math> (the inverse trigonometric function) as the part to differentiate and <math>1/x</math> (the algebraic function) as the part to integrate. We get:

			<math>\int \frac{\arctan x}{x} \, dx = (\arctan x)(\ln x) - \int \frac{\ln x \, dx}{x^2 + 1}</math>

			Now, ILATE suggests that for the integral on the right side, we take <math>\ln x</math> (the logarithmic function) as the part to differentiate) and <math>1/(x^2 + 1)</math> as the part to integrate. ''However'', we immediately see that making this choice would fall into the circular trap.

			In fact, there is no other reasonable choice to make, so this integral cannot be done using the standard function integrations. Note, however, that we ''did'' manage to establish an equivalence of integration problems:

			<math>\int \frac{\arctan x}{x} \, dx = \int \frac{\ln x}{x^2 + 1} \, dx</math>

Vipul: Created page with "==Circular trap== When using integration by parts a second time, make sure you don't choose as the ''part to integrate'' the thing you got by differentiating the ''part to di..."

2012-04-10T19:09:59Z

Created page with "==Circular trap== When using integration by parts a second time, make sure you don't choose as the ''part to integrate'' the thing you got by differentiating the ''part to di..."

New page

==Circular trap==

When using integration by parts a second time, make sure you don't choose as the ''part to integrate'' the thing you got by differentiating the ''part to differentiate'' from the original product. Otherwise, you get in a circular trap and don't get any new information. The most typical application of integration by parts a second time is if you choose to differentiate ''again'' the expression that you already obtained through differentiation the first time.

==Formal explanation of circular trap==

===Explanation using function notation===

Consider an integration of the form:

<math>\int F(x) g(x) \, dx</math>

where <math>G</math> is an antiderivative for <math>g</math>. Then:

<math>\int F(x) g(x) \, dx = F(x)G(x) - \int F'(x)G(x) \, dx</math>

Suppose that, for the new integral, we choose <math>G</math> as the part to differentiate and <math>F'</math> as the part to integrate. The expression then simplifies to:

<math>\int F(x) g(x) \, dx = F(x)G(x) - [G(x)F(x) - \int g(x)F(x) \, dx]</math>

Simplifying, we get:

<math>\int F(x) g(x) \, dx = \int F(x)g(x) \, dx</math>

In other words, ''we ended up with the original expression, and obtained no new information in the process''. Another way of saying this is that we went in circles, i.e., went back along the path we came from, hence did not make any progress.

===Explanation using dependent-independent variable notation===

Consider an integration of the form:

<math>\int u \frac{dv}{dx} \, dx</math>

Using integration by parts, write this as:

<math>\int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx</math>

Now, let's say that, for the right side integral, we pick our "new" <math>u</math> to be <math>v</math> and our "new" <math>dv/dx</math> to be <math>du/dx</math>. Then, our "new" <math>v</math> is the "old" <math>u</math> and we get:

<math>\int u \frac{dv}{dx} \, dx = uv - [vu - \int u \frac{dv}{dx} \, dx]</math>

Simplifying, we get:

<math>\int u \frac{dv}{dx} \, dx = \int u \frac{dv}{dx} \, dx</math>

In other words, ''we ended up with the original expression, and obtained no new information in the process''. Another way of saying this is that we went in circles, i.e., went back along the path we came from, hence did not make any progress.