Can You Calculate Factorials for Non-Integer Numbers Using Integration by Parts?

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Discussion Overview

The discussion revolves around the application of integration by parts, particularly in the context of calculating integrals involving exponential and trigonometric functions. Participants explore the technique and its implications, including the potential for calculating factorials of non-integer numbers through integration methods.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant presents an integral involving \( e^{2x} \) and \( \cos(3x) \) and seeks guidance on how to proceed after an initial application of integration by parts.
  • Another participant provides a detailed breakdown of the integration by parts process, suggesting a method to solve the integral by applying the technique multiple times.
  • A third participant emphasizes the importance of recognizing when to apply integration by parts repeatedly, noting that this is a common occurrence in calculus.
  • Further, a participant mentions a connection between integration by parts and the calculation of factorials for all positive real numbers, suggesting that integrating \( e^{-x} x^n \) leads to \( n! \) through repeated applications of the technique.
  • Another participant raises a question about the behavior of the function as it approaches zero at certain limits, indicating a curiosity about the mathematical properties involved.

Areas of Agreement / Disagreement

Participants generally agree on the utility of integration by parts and its application to the problem at hand. However, there are varying levels of understanding and approaches to the problem, with no consensus on the best method for calculating the factorial of non-integer numbers or the specifics of the integral's behavior.

Contextual Notes

Some participants express uncertainty regarding the limits of integration and the behavior of functions at infinity, indicating that assumptions about convergence or the applicability of certain methods may not be fully resolved.

lord12
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integral of (e^(2x))(cos3x)dx

I get 1/2e^(2x)sin2x - 3/2integral(e^(2x)cos3xdx)

what do i do next?
 
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Let [tex]u = e^{2x}[/tex] [tex]dv = \cos 3x[/tex] [tex]du = 2e^{2x}[/tex] [tex]v = \frac{1}{3}\sin 3x[/tex]Since [tex]\int udv = uv-\int vdu[/tex], we have:

[tex]\frac{e^{2x}}{3}\sin 3x - \frac{2}{3}\int e^{2x}\sin 3x = \int e^{2x}\cos 3x \; dx[/tex]

Now apply integration by parts to [tex]\frac{2}{3}\int e^{2x}\sin 3x[/tex] and you can solve for the integral.
 
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Please be more careful with what you write!
[tex]I=\int{e^{2x}\cos(3x)dx=\frac{1}{2}e^{2x}\cos(3x)+\frac{3}{2}\int{e}^{2x}\sin(3x)dx+C[/tex]
where C is some integration constant.

Now, the right-hand side indefinite integral may be rewritten as:
[tex]\int{e}^{2x}\sin(3x)dx=\frac{1}{2}e^{2x}\sin(3x)-\frac{3}{2}\int{e}^{2x}\cos(3x)dx=\frac{1}{2}e^{2x}\sin(3x)-\frac{3}{2}I[/tex]
And your first expression may then be written as:

[tex]I=\frac{1}{2}e^{2x}\cos(3x)+\frac{3}{2}(\frac{1}{2}e^{2x}\sin(3x)-\frac{3}{2}I)+C[/tex]

Now, solve for I.
 
I think lord12s main problem was that he has never encountered a situation where he has to apply integration by parts twice. It occurs often, and many times more than just once. In fact many go on an infinite number of applications.
 
Gib Z said:
It occurs often, and many times more than just once. In fact many go on an infinite number of applications.

It would definitely be interesting to see what a man who applied integration by parts for an infinite number of times would look like after he did so. :biggrin: :-p
 
lol yea that would be. But in case your looking for one where integration is applied [itex]n[/itex] times, and [itex]n[/itex] can be as large as you want, then look at Euler's ...Totient Function? I can't remember the name, but it basically stated:

[tex]\int e^{-x} x^n = n![/tex]

I meant to write, when integrated from infinity to 0, but don't know how..

anyway its proven by applying Integration by parts n times, try it. Btw this leads us to be able to calculate factorials all positive real numbers, not just integers. Its very tedious to do approximate by hand, but in case you want to, just to impress someone that you can find the factorial of 5.6, then use the trapezoidal rule, much simpler and accurate than simpson's rule in this case. And approximate by finding area of 0 to 4n, infinity may be a bit hard...I don't know why the function seems to rapidly dive down to zero at 4n, maybe someone can answer that for me?
 
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