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

In summary, the integral of (e^(2x))(cos3x)dx can be solved using integration by parts. After applying the rule, the expression can be rewritten and solved for the integral. This method can be applied multiple times, even to an infinite number of applications, as seen in Euler's Totient Function. The trapezoidal rule can also be used to approximate the factorial of non-integer numbers.
  • #1
lord12
9
0
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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  • #2
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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  • #3
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.
 
  • #4
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.
 
  • #5
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
 
  • #6
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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Related to Can You Calculate Factorials for Non-Integer Numbers Using Integration by Parts?

1. What is integration by parts?

Integration by parts is a mathematical method used to find the integral of a product of two functions. It is based on the product rule of differentiation and is particularly useful for solving integrals involving polynomials, exponential functions, and trigonometric functions.

2. When do I use integration by parts?

Integration by parts is typically used when the integral involves a product of two functions that cannot be easily simplified or evaluated using other methods, such as substitution or partial fractions. It is also useful for solving integrals that involve logarithmic functions or inverse trigonometric functions.

3. What is the formula for integration by parts?

The formula for integration by parts is ∫udv = uv - ∫vdu, where u and v are the two functions being multiplied together and du and dv are their respective differentials. This formula is derived from the product rule of differentiation.

4. How do I choose u and dv for integration by parts?

When using integration by parts, u and dv are chosen based on a specific order of preference known as the "ILATE" rule. "ILATE" stands for Inverse trigonometric functions, Logarithmic functions, Algebraic functions, Trigonometric functions, and Exponential functions. The function that falls first in this order is chosen as u, while the other function is chosen as dv.

5. Are there any special cases for integration by parts?

Yes, there are a few special cases for integration by parts. One is when the integrand contains a polynomial multiplied by a power of x, in which case u can be chosen as the polynomial and dv as the power of x. Another special case is when the integrand contains a logarithmic function multiplied by a power of x, in which case u can be chosen as the logarithmic function and dv as the power of x.

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