- #1
quicksilver123
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Difficulties with Substitution Rule (integration)
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The discussion centers on the challenges of understanding the substitution rule in integration, particularly its relationship to the chain rule. A key point is the distinction between the substitution rule and the Leibniz rule, with the former corresponding to the chain rule and the latter to integration by parts. The correct application of the substitution rule involves recognizing that the integral of a product cannot be separated into individual integrals. The conversation highlights the importance of proper notation and understanding how to manipulate integrals using Leibniz notation for clarity. Ultimately, mastering these concepts is crucial for effectively solving indefinite integrals.
- #2
quicksilver123
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The Lamar university web page on integration offers the same material and proof.
- #3
- 20,760
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If ##g(x)=v## then ##\int [f(g(x))g'(x)]dx= \int [f(v)v'(x)]dx=\int [f(v)]dv = \int f(u)du##.
Your mistake is ##\int [f(v)v'(x)]dx \neq [\int f(v)dv] \cdot [\int v'dv]##.
The essential reason for this is, because a differentiation obeys the Leibniz rule, i.e. leads to a derivative: ##d(f\cdot g)= (df)\cdot g + f\cdot (dg),## and does not allow a factor-wise multiplication: ##d(f\cdot g)\neq (df)\cdot (dg)\,.##
Your mistake is ##\int [f(v)v'(x)]dx \neq [\int f(v)dv] \cdot [\int v'dv]##.
The essential reason for this is, because a differentiation obeys the Leibniz rule, i.e. leads to a derivative: ##d(f\cdot g)= (df)\cdot g + f\cdot (dg),## and does not allow a factor-wise multiplication: ##d(f\cdot g)\neq (df)\cdot (dg)\,.##
- #4
quicksilver123
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Could you explicitly explain the correct method in terms of your liebnitz Rule?
- #5
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Sorry, I confused the chain rule and the Leibniz rule. The chain rule corresponds to the substitution rule and the Leibniz rule corresponds to integration by parts. The shortest way to see the equation in (4) is to use the notation with the ##d##'s, also called Leibniz notation. Here we get by the substitution ##g(x)=u##quicksilver123 said:
$$
\int [f(g(x))g'(x)dx] = \int [f(g(x)) \frac{dg(x)}{dx} dx] = \int [f(g(x))dg(x)] = \int [f(u)du]
$$
where I have only calculated with the terms under the integration. Nothing goes outside as I understood your remark.
You haven't written (in your hand writing) what your integration variable is, so I assume you meant ##\int [f(g(x))g'(x)]dx = [\int f(g(x))dx] \cdot [\int u'dx] =[\int f(g(x))dx] \cdot u##. But the integral of a product (and that is why I brought up the Leibniz rule) is given by
$$
[f(g(x))\cdot g(x)]' = f(g(x)) \cdot g'(x) + [f(g(x)]' \cdot g(x) \Longrightarrow \int [f(g(x)) \cdot g'(x)]dx = f(g(x))\cdot g(x) - \int [f(g(x))]' \cdot g(x)dx
$$
First, I tried to show that ##f_n## converges uniformly on ##[0,2\pi]##, which is true since ##f_n \rightarrow 0## for ##n \rightarrow \infty## and ##\sigma_n=\mathrm{sup}\left| \frac{\sin\left(\frac{n^2}{n+\frac 15}x\right)}{n^{x^2-3x+3}} \right| \leq \frac{1}{|n^{x^2-3x+3}|} \leq \frac{1}{n^{\frac 34}}\rightarrow 0##. I can't use neither Leibnitz's test nor Abel's test. For Dirichlet's test I would need to show, that ##\sin\left(\frac{n^2}{n+\frac 15}x \right)## has partialy bounded sums...
It's easy to see that any polynomial is function of that class. Also it seems that composition of exponent and polynomial is good as well. G_f(a, b) should be equal G_f(b, a) as the f(a+b) is symmetrical.
Does anyone have information about this or at least related to it?