Complex Integration by Parts Help. Please

In summary, the person is trying to integrate an integer function by parts and is stuck. They suggest integrating by parts twice and getting back to the original integral.
  • #1
Silmax
4
0
Hi
Can anyone help me with this integration.
I will use I to symbol the integer sign

Limits between 1 and 0

Ie^-x.Sinxdx

I understand I have to integrate by parts and get the following answer, ignoring the limits for the time being.

Ie^-x.Sinx = e^-x.Cosx I -Cosx.e^-x dx

then i believe i have to integrate the second part of the sum ( I -Cosx.e^-xdx)and i get.

I -Cosx.e^-xdx= -Cosx.-e^-x- I -e^-x.Sinx

Now i am stuck.
How do I combine the two answers to get a single sum which i can then put between limits.??

Thank you.
 
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  • #2
Is this what you are trying to integrate:

[tex]I=\int_{0}^{1}e^{-x}sin(x)dx...?[/tex]

I don't see any complex integral here. I might be wrong as well, because my eyes are going...lol...

let [tex] u=e^{-x}, du=-e^{-x}dx,and,v=\int sin(x)dx=-cosx[/tex]

Integrate by parts twice, and you will get sth in terms of the original integral, and you will be fine.
 
  • #3
The OP might be referring to the integral

[tex]\int^{1}_{0} ie^{-x} \sin (x) dx[/tex]

?
 
  • #4
[tex]S=\int_{0}^{1}e^{-x}sinxdx.[/tex]
This is the actual sum. Now I have learned how to set it up properly.
Thanks
 
  • #5
sutupidmath said:
Is this what you are trying to integrate:

[tex]I=\int_{0}^{1}e^{-x}sin(x)dx...?[/tex]

I don't see any complex integral here. I might be wrong as well, because my eyes are going...lol...

let [tex] u=e^{-x}, du=-e^{-x}dx,and,v=\int sin(x)dx=-cosx[/tex]

Integrate by parts twice, and you will get sth in terms of the original integral, and you will be fine.

There is no need for subsitution here - actually I don't see how it helps, you'd get something like [itex]\sin\log u[/itex]...not really helpful.

Just use integration by parts right away, and you'll get back to the integral you started from for which you can then solve.
 
  • #6
Pere Callahan said:
There is no need for subsitution here - actually I don't see how it helps, you'd get something like [itex]\sin\log u[/itex]...not really helpful.

Just use integration by parts right away, and you'll get back to the integral you started from for which you can then solve.

I don't see myself saying to use substitution lol...!

And as long as i can see, i also suggested integration by parts, and that twice.

We might have different concepts of what substitution and integration by parts is, who knows!

OR... Probbably my eyes are really going lol...
 
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  • #7
Oh true you didn't mention substitution explicitly, my fault.

But why did write down then what du is ...hm, doesn't matter. I guess our concepts of subsitution and integration by parts are rather similar.:smile:
 
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  • #8
Pere Callahan said:
Oh true you didn't mention substitution explicitly

Neither explicitly nor implicitly!
 

1. What is complex integration by parts?

Complex integration by parts is a method used in calculus to evaluate integrals involving products of functions. It is similar to the integration by parts method for real integrals, but it involves using complex numbers and complex-valued functions.

2. When should complex integration by parts be used?

Complex integration by parts should be used when the integral involves a product of complex-valued functions. It is also useful when the integral cannot be evaluated using other methods, such as substitution or partial fractions.

3. How do you use the complex integration by parts method?

To use the complex integration by parts method, you must first identify the function to be integrated and the function to be differentiated. Then, you can use the following formula: ∫u(x)v'(x)dx = u(x)v(x) - ∫u'(x)v(x)dx, where u(x) and v(x) are the functions chosen for integration and differentiation, respectively.

4. Can complex integration by parts be used for definite integrals?

Yes, complex integration by parts can be used for definite integrals. After using the integration by parts formula, you can substitute the limits of integration to obtain the final result.

5. Are there any tips for solving complex integration by parts problems?

One tip for solving complex integration by parts problems is to choose u(x) and v(x) carefully. It is usually easier to choose u(x) as the function that becomes simpler after differentiating, and v(x) as the function that becomes more complex after integration. It is also important to practice and familiarize yourself with the method to become more efficient at solving problems.

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