Creative integration by parts

In summary: Not quite. You were off by a negative sign and you had a sine in the integral instead of cosine.I forgot to evaluate the left hand side. at infinity, the exponential goes to zero. at zero, the trig term goes to zero. So that's "done".You mean the left term on the RHS right?:wink:...the LHS of the equation is just dI/dx!Okay, so now you should have:\frac{\partial I}{\partial x}=-\frac{x}{2kt}\int_{0}^{\infty}e^{-k\omega^{2}t}\cos(\
  • #1
paxprobellum
25
0
"Creative" integration by parts

Homework Statement


Evaluate [tex]I=\int_{0}^{Inf}e^{-kw^{2}t}cos(wx)dw[/tex] in the following way. Determine (partial derivative) dI/dx, then integrate by parts.

Homework Equations



[tex]\int udV = uv - \int vdU[/tex]


The Attempt at a Solution



So I have to figure out dI/dx first. I think it should be a function that lends hand to integration by parts. Obviously it has a trig portion and possibly an exponential/algebraic component. I'm a little sketchy on the details of extracting dI/dx from that equation though. My guess is:

dI/dx = (1/x)*coswx

so
[tex]I = \int (1/x)*cos(wx)dx[/tex]
Integration by parts [ u=1/x, dv = cos(wx)dx ] and I get:

[tex] (1/xw)sin(wx)+(1/w) \int (1/x^2)sin(wx)dx [/tex]

which doesn't seem to be heading in the right direction. Any thoughts?
 
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  • #2


paxprobellum said:
My guess is:

dI/dx = (1/x)*coswx

Huh?! Where the heck did this "guess" come from? :confused:...there is no need to guess at all.

If [tex]I=\int_{0}^{\infty}e^{-k\omega^{2}t}\cos(\omega x)d\omega[/tex]

Then you should clearly see that:

[tex]\frac{\partial I}{\partial x}=\frac{\partial}{\partial x} \int_{0}^{\infty}e^{-k\omega^{2}t}\cos(\omega x)d\omega[/tex]

Right?:wink:

Hint: If [itex]\omega[/itex] and [itex]x[/itex] are independent, does it matter whether you first integrate over [itex]\omega[/itex] and then differentiate with respect to [itex]x[/itex] or whether you first differentiate with respect to [itex]x[/itex] and then integrate over [itex]\omega[/itex]?:wink:
 
  • #3


gabbagabbahey said:
Huh?! Where the heck did this "guess" come from? :confused:...there is no need to guess at all.

If [tex]I=\int_{0}^{\infty}e^{-k\omega^{2}t}\cos(\omega x)d\omega[/tex]

Then you should clearly see that:

[tex]\frac{\partial I}{\partial x}=\frac{\partial}{\partial x} \int_{0}^{\infty}e^{-k\omega^{2}t}\cos(\omega x)d\omega[/tex]

Right?:wink:

Hint: If [itex]\omega[/itex] and [itex]x[/itex] are independent, does it matter whether you first integrate over [itex]\omega[/itex] and then differentiate with respect to [itex]x[/itex] or whether you first differentiate with respect to [itex]x[/itex] and then integrate over [itex]\omega[/itex]?:wink:

Yer I guess that wasn't a very good guess, mmm? :P

So exchanging the order of operations, I get:
[tex]

\int_{0}^{infty} \frac{\partial}{\partial x} e^{-kw^{2}t}*cos(wx)dw [/tex]

giving:
[tex]- \int w*sin(wx)*e^{-kw^{2}t}dw[/tex]

This seems even worse than before. Any ideas from here?
 
Last edited:
  • #4


paxprobellum said:
Yer I guess that wasn't a very good guess, mmm? :P

So exchanging the order of operations, I get:
[tex]

\int_{0}^{infty} \frac{\partial}{\partial x} e^{-kw^{2}t}*cos(wx)dw [/tex]

giving:
[tex]- \int w*sinx*e^{-kw^{2}t}dw[/tex]

Isn't [tex]\frac{\partial}{\partial x}\cos(\omega x)=-\omega \sin(\omega x)[/tex] ?:wink:
 
  • #5


gabbagabbahey said:
Isn't [tex]\frac{\partial}{\partial x}\cos(\omega x)=-\omega \sin(\omega x)[/tex] ?:wink:

Yeah sorry. I caught that after I posted. See above.
 
  • #6


paxprobellum said:
[tex]- \int w*sin(wx)*e^{-kw^{2}t}dw[/tex]

This seems even worse than before. Any ideas from here?

Try using the hint they gave you and integrate by parts...I recommend using [itex]u=\sin(\omega x)[/itex] and [itex]dv=-\omega e^{-k\omega^2 t}d\omega[/itex] :wink:
 
  • #7


gabbagabbahey said:
Try using the hint they gave you and integrate by parts...I recommend using [itex]u=\sin(\omega x)[/itex] and [itex]dv=-\omega e^{-k\omega^2 t}d\omega[/itex] :wink:

I seem to be stuck in a loop. Integration by parts gives me

[tex]
-(1/2kt)*[sin(w*x)*e^{-kw^{2}t} + x \int e^{-kw^{2}t} sin(wx)dw ]
[/tex]

to do that integral, I'd need to do it by parts. Using the same strategy, I'd end up with a very similar integral. Except this time, I can't do the dV analytically. Or at least I don't think I could do [tex] \int e^{-kw^{2}t}dw [/tex]
 
  • #8


paxprobellum said:
I seem to be stuck in a loop. Integration by parts gives me

[tex]
-(1/2kt)*[sin(w*x)*e^{-kw^{2}t} + x \int e^{-kw^{2}t} sin(wx)dw ]
[/tex]

That doesn't look right...you should be getting

[tex]\frac{\partial I}{\partial x}=\frac{\sin(\omega x)e^{-k\omega^{2}t}}{2kt}{\left|_{\omega=0}^{\infty}}-\frac{x}{2kt}\int_{0}^{\infty}e^{-k\omega^{2}t}\cos(\omega x)d\omega[/tex]Once you correct your mistakes and get the above result, evaluate the first term on the right-hand side by substituting in the limits...what does that leave you with?:wink:
 
Last edited:
  • #9


gabbagabbahey said:
That doesn't look right...you should be getting

[tex]\frac{\partial I}{\partial x}=\frac{\sin(\omega x)e^{-k\omega^{2}t}}{2kt}\left|_{\omega=0}^{\infty}-\frac{x}{2kt}\int_{0}^{\infty}e^{-k\omega^{2}t}\cos(\omega x)d\omega[/tex]Once you correct your mistakes and get the above result, evaluate the first term on the right-hand side by substituting in the limits...what does that leave you with?:wink:

I'm confused. What you wrote above is the exact same as what I wrote (i pulled the -1/2kt out of the whole thing).

I forgot to evaluate the left hand side. at infinity, the exponential goes to zero. at zero, the trig term goes to zero. So that's "done".

I'm not sure what you meant about the RHS. I could do it by parts, but it doesn't seem to help. It just gets uglier.
 
  • #10


paxprobellum said:
I'm confused. What you wrote above is the exact same as what I wrote (i pulled the -1/2kt out of the whole thing).

Not quite. You were off by a negative sign and you had a sine in the integral instead of cosine.
I forgot to evaluate the left hand side. at infinity, the exponential goes to zero. at zero, the trig term goes to zero. So that's "done".

You mean the left term on the RHS right?:wink:...the LHS of the equation is just dI/dx!

Okay, so now you should have:

[tex]\frac{\partial I}{\partial x}=-\frac{x}{2kt}\int_{0}^{\infty}e^{-k\omega^{2}t}\cos(\omega x)d\omega[/tex]

But wait! Isn't that integral exactly the same as your original integral 'I' ? :smile:

[tex]\implies \frac{\partial I(x,t)}{\partial x}=-\frac{x}{2kt} I(x,t)[/tex]

Which is a separable PDE for [itex]I(x,t)[/itex]...you should know how to solve that one (don't forget to include a constant (a function that does not depend on x, but may depend on t) of integration!)...
 
  • #11


gabbagabbahey said:
Not quite. You were off by a negative sign and you had a sine in the integral instead of cosine.

Whoops. I guess its getting late? :P

gabbagabbahey said:
You mean the left term on the RHS right?:wink:...the LHS of the equation is just dI/dx!

Oh, right. I understand what you mean. I was just referring to the LHS/RHS of the right side.

gabbagabbahey said:
Okay, so now you should have:

[tex]\frac{\partial I}{\partial x}=-\frac{x}{2kt}\int_{0}^{\infty}e^{-k\omega^{2}t}\cos(\omega x)d\omega[/tex]

But wait! Isn't that integral exactly the same as your original integral 'I' ? :smile:

Oh very clever! Crazy math tricks!

Thanks a bunch.
 
  • #12


After you've solved the PDE, you can determine the constant of integration by evaluating your original integral for x=0 and using the result.
 
  • #13


gabbagabbahey said:
After you've solved the PDE, you can determine the constant of integration by evaluating your original integral for x=0 and using the result.

Excellent idea! I didn't think the constant could be solved for. I did:

[tex]
I(0,t) = \int _{0}^{Inf} e^{-kw^{2}t} dw = \frac {1}{2} [\frac {\pi}{kt}] ^{(1/2)} = C

[/tex]

Thanks again for your help. I am struggling on this problem set for some reason (see my other "talk to myself" post :P). Almost done! :)
 

Related to Creative integration by parts

1. What is creative integration by parts?

Creative integration by parts is a mathematical technique used to evaluate integrals that involve products of functions. It involves breaking down the integral into smaller parts and using the product rule to integrate one part while differentiating the other. This method is useful for solving more complex integrals and can be applied in various fields such as physics, engineering, and economics.

2. How does creative integration by parts differ from regular integration by parts?

Creative integration by parts is a more flexible approach compared to regular integration by parts. It allows for more creativity in choosing the functions to integrate and differentiate, whereas regular integration by parts follows a specific formula. This makes creative integration by parts more useful for solving complex integrals that cannot be easily evaluated using the traditional method.

3. What are some examples of when creative integration by parts is useful?

Creative integration by parts can be used to solve integrals involving trigonometric functions, logarithmic functions, and exponential functions. It is also useful for evaluating integrals with unconventional forms, such as those with a fraction in the exponent or those with multiple variables.

4. Are there any limitations to using creative integration by parts?

While creative integration by parts is a powerful technique, it may not always be the most efficient method for evaluating integrals. It requires a good understanding of the product rule and may involve multiple steps, making it time-consuming for some integrals. In some cases, other methods such as substitution or partial fraction decomposition may be more suitable.

5. How can I improve my skills in using creative integration by parts?

Practice is key to improving your skills in creative integration by parts. Start with simpler integrals and gradually work your way up to more complex ones. It is also helpful to have a good understanding of the basic integration rules and techniques. Additionally, seeking guidance from a tutor or studying worked examples can also improve your proficiency in using this method.

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