Creative integration by parts

[paxprobellum]

"Creative" integration by parts

Homework Statement

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.

Homework Equations

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

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?

 

[gabbagabbahey]

My guess is:

dI/dx = (1/x)*coswx

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 $\omega$?

 

[paxprobellum]

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 $\omega$?

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?

 

[gabbagabbahey]

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:
$$- \int w*sinx*e^{-kw^{2}t}dw$$

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

 

[paxprobellum]

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

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

 

[gabbagabbahey]

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

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

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

 

[paxprobellum]

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

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

 

[gabbagabbahey]

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

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

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?

 

[paxprobellum]

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?

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.

 

[gabbagabbahey]

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?...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!)...

 

[paxprobellum]

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

You mean the left term on the RHS right?...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.

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' ?

Oh very clever! Crazy math tricks!

Thanks a bunch.

 

[gabbagabbahey]

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.

 

[paxprobellum]

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:

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

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! :)