Integration by parts homework help

In summary, the conversation involved the speaker trying to solve an integral of the form \int_0^{infinity} \ e^{-s*t}*t*cos(t) dt. They first attempted integration by parts with u=t*cos(t) and dv=e^(-s*t), but were unable to make progress. They then tried using the rule \mathcal{L}\{t^n g(t)\}=(-1)^n\frac{d^n}{ds^n}\mathcal{L}\{g(t)\} but were still unable to solve the integral. The speaker also mentioned that they got an indeterminate form when using L'Hopital's rule. The expert summarizer requested to see the work to determine where the speaker went wrong
  • #1
Dustinsfl
2,281
5
[tex]\int_0^{infinity} \ e^{-s*t}*t*cos(t) dt[/tex]

I tried integration by parts with u=t*cost and dv=e^(-s*t) but that didn't get anywhere.

I then tried: [tex]\L{t^n*g(t)}=(-1)^n d/ds[\int_0^{infinity} \ e^{-s*t}*cos(t) dt[/tex] but again nothing was working.

This is a Laplace Transformation where ft=t cos(t)
 
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  • #2


Dustinsfl said:
I then tried: [tex]\L{t^n*g(t)}=(-1)^n d/ds[\int_0^{infinity} \ e^{-s*t}*cos(t) dt[/tex] but again nothing was working.

Do you mean that you tried applying the rule

[tex]\mathcal{L}\{t^n g(t)\}=(-1)^n\frac{d^n}{ds^n}\mathcal{L}\{g(t)\}[/tex]

If so, that should be fairly easy, provided that you know what the Laplace transform of [itex]\cos(t)[/itex] is.
 
  • #3


The problem was I get cos(t)/(s*e^(-st)) which is in determinant and using l'hopitals rule didn't help.

Also, the integral after separation by parts is now e time sin then you get an in determinant form with sin.
 
  • #4


Dustinsfl said:
The problem was I get cos(t)/(s*e^(-st)) which is in determinant and using l'hopitals rule didn't help.

How on Earth are you getting something that depends on [itex]t[/itex] after integrating over [itex]t[/itex]? :confused:

Show your work so I can see where you are going wrong.
 
  • #5


Work

Both methods attempted

Never mind the image won't load.
 

What is Integration by Parts?

Integration by Parts is a method used to solve integrals that involve the product of two functions. It is based on the product rule of differentiation and is useful for integrating functions that cannot be easily integrated using other methods.

When is Integration by Parts used?

Integration by Parts is used when the integrand (the function being integrated) is a product of two functions, and one of the functions cannot be easily integrated by other methods. It is also useful when the integrand involves a polynomial multiplied by a transcendental function such as exponential, logarithmic, or trigonometric functions.

What is the formula for Integration by Parts?

The formula for Integration by Parts is ∫udv = uv - ∫vdu, where u and v are functions and dv is the differential of v. This formula is derived from the product rule of differentiation, which states that d(uv)/dx = u(dv/dx) + v(du/dx).

Can Integration by Parts be used to solve definite integrals?

Yes, Integration by Parts can be used to solve both indefinite and definite integrals. For definite integrals, the limits of integration must be substituted into the final result after integration by parts has been applied.

What are some tips for using Integration by Parts effectively?

Some tips for using Integration by Parts effectively include choosing u and dv carefully, using the LIPET rule (Logarithmic, Inverse trigonometric, Polynomial, Exponential, Trigonometric) to determine u and dv, and using algebraic manipulation or trigonometric identities to simplify the integrand before applying Integration by Parts.

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