# Creative integration by parts

1. Feb 17, 2009

### paxprobellum

"Creative" integration by parts

1. The problem statement, all variables and given/known data
Evaluate $$I=\int_{0}^{Inf}e^{-kw^{2}t}cos(wx)dw$$ in the following way. Determine (partial derivative) dI/dx, then integrate by parts.

2. Relevant equations

$$\int udV = uv - \int vdU$$

3. 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
$$I = \int (1/x)*cos(wx)dx$$
Integration by parts [ u=1/x, dv = cos(wx)dx ] and I get:

$$(1/xw)sin(wx)+(1/w) \int (1/x^2)sin(wx)dx$$

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

2. Feb 17, 2009

### gabbagabbahey

Re: "Creative" integration by parts

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

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

Then you should clearly see that:

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

Right?

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

3. Feb 17, 2009

### paxprobellum

Re: "Creative" integration by parts

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

So exchanging the order of operations, I get:
$$\int_{0}^{infty} \frac{\partial}{\partial x} e^{-kw^{2}t}*cos(wx)dw$$

giving:
$$- \int w*sin(wx)*e^{-kw^{2}t}dw$$

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

Last edited: Feb 17, 2009
4. Feb 17, 2009

### gabbagabbahey

Re: "Creative" integration by parts

Isn't $$\frac{\partial}{\partial x}\cos(\omega x)=-\omega \sin(\omega x)$$ ???

5. Feb 17, 2009

### paxprobellum

Re: "Creative" integration by parts

Yeah sorry. I caught that after I posted. See above.

6. Feb 17, 2009

### gabbagabbahey

Re: "Creative" integration by parts

Try using the hint they gave you and integrate by parts....I recommend using $u=\sin(\omega x)$ and $dv=-\omega e^{-k\omega^2 t}d\omega$

7. Feb 17, 2009

### paxprobellum

Re: "Creative" integration by parts

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

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

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 $$\int e^{-kw^{2}t}dw$$

8. Feb 17, 2009

### gabbagabbahey

Re: "Creative" integration by parts

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

$$\frac{\partial I}{\partial x}=\frac{\sin(\omega x)e^{-k\omega^{2}t}}{2kt}{\left|_{\omega=0}^{\infty}}-\frac{x}{2kt}\int_{0}^{\infty}e^{-k\omega^{2}t}\cos(\omega x)d\omega$$

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?

Last edited: Feb 17, 2009
9. Feb 17, 2009

### paxprobellum

Re: "Creative" integration by parts

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. Feb 17, 2009

### gabbagabbahey

Re: "Creative" integration by parts

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

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

Okay, so now you should have:

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

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

$$\implies \frac{\partial I(x,t)}{\partial x}=-\frac{x}{2kt} I(x,t)$$

Which is a separable PDE for $I(x,t)$....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. Feb 17, 2009

### paxprobellum

Re: "Creative" integration by parts

Whoops. I guess its getting late? :P

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

Oh very clever! Crazy math tricks!

Thanks a bunch.

12. Feb 17, 2009

### gabbagabbahey

Re: "Creative" integration by parts

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. Feb 17, 2009

### paxprobellum

Re: "Creative" integration by parts

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

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

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! :)