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jcain6
September 5, 2005, 05:18 PM
I am doing some physics homework and I have come across three integrals that are giving me problems.


1) dx/(a^2+x^2)^2 from negative infinity to infinity

2) (x^2 dx)/(a^2+x^2)^2 from negative infinity to infinity

3) (sinkx)^2/x^2 from 0 to infinity

I would appreciate any help that anyone has to offer. #3 resembles the Dirichlet Integral (if we drop off the squares) and then it would be pi/2. All k's and a's are just constants.

Thanks,

jcain6

EricK
September 5, 2005, 06:24 PM
My integration is a little rusty. But here are some suggestions (don't know if they are any good)

For 1, you might want to try some substitution like x= a tan(y)

For 2, maybe do partial fractions and treat as in 1

For 3, does integration by parts work?

Eric

jcain6
September 5, 2005, 09:07 PM
Thanks for the suggestions, I hadn't thought of the x=atany substitution - unfortunately I couldn't get that to work (almost though). On the third integral when I do try integration by parts I get an even more complex integral that doesn't seem to do any good nor does it repeat itself. Thanks for the suggestions.

starling
September 5, 2005, 09:24 PM
That's what I don't like about differential equations. It's like: you have the problem, but there is literally no way to reach the solution, beyond blind luck, or external knowledge. Some special Integrals have general solutions, but not all of them. You can't, as Euclid did, start with a few simple principles, and deduce all of Algebra only based on those principles. To be honest, what you have to do is know with your homework is the solution before you solve it.

The good news is your math book is going to give you the false hope that all problems have known solutions, and so all you have to do is look up which solution they want you to get, and start plugging things in. The bad news is you won't really learn how they got that solution in the first place. So...

Since there's no 'da' in 1 or 2, you can consider 'a' like a constant. Try finding something that when you take the derivative, you get A/(B+x^2)^2. Just guess, use your best judgement, check the chapter for example problems that match that. That's the only thing you'll be taught to do: solve problems depending on on previous "magically" solved examples.

Notice that (2) is the same as (1), except it's A*x^2/(B+x^2)^2. As for (3), I don't even know what the sinkx is. It surely isn't a synx! Did you mean sin(kx)? Go look for your trigonometric equivalencies and try to find stuff equal to the sine squared. Just fiddle around randomly, trying to get your problem to match a previously solved example. Once you do that, then you can just emulate the example, just like every professional mathemetician does. I think some people even spent 20 years finding a solution to one of those problems that didn't have a solution discovered already.

Sometimes I think mathemeticians must have invented time travel, so they could go back in time and give their ancestors the solutions to differential equations. That's the only way I can fathom people figured out what the integral(dx/(cos(x)+sin(x)) is!

repoman
September 5, 2005, 11:18 PM
I think that your 3rd problem needs contour integreation, but I am not good at that at all.

uncool
September 5, 2005, 11:32 PM
The first one should work via x = a*tan(y)
It becomes a*sec^2(y)/(a^2*sec^2(y))dy = dy/a
Integrate, get y/a + C, y = arctan(x/a), so arctan(x/a)/a

Second one should work as well. You get tan^2(y)/a dy = sec^2(y)/a - 1/a dy
integrate, get tan(y)/a - y/a + C = x/a^2 - arctan(x/a)/a + C

Third one, you get an unintegrable thing. e^x/x is not integrable, and you get an integral of that form if you try to integrate.
-Uncool-

repoman
September 5, 2005, 11:53 PM
3rd problem:

(Sin(kx))^2 = 1/2*(1-cos(2kx))

Then let u = kx^2
du = 2kx dx
dx = (1/2)*(kx)^-0.5 du

might work

Stu
September 6, 2005, 01:34 AM
I'm getting something awfully close to k*pi/2 for the third one using numerical methods, but I'm not too sure about an analytical solution for that.

EricK
September 6, 2005, 02:01 PM
For the third one, if you use Integration by parts:
u = sin^2(kx) v' = 1/x^2 (so u' = 2ksin(kx)cos(kx) = ksin(2kx), v = -1/x)
You end up with a doable integral, I think

Eric

Barbarian
September 6, 2005, 02:27 PM
I am doing some physics homework and I have come across three integrals that are giving me problems.


1) dx/(a^2+x^2)^2 from negative infinity to infinity

2) (x^2 dx)/(a^2+x^2)^2 from negative infinity to infinity

3) (sinkx)^2/x^2 from 0 to infinity

I would appreciate any help that anyone has to offer. #3 resembles the Dirichlet Integral (if we drop off the squares) and then it would be pi/2. All k's and a's are just constants.

Thanks,

jcain6
Obviously you are looking for tips on how to do these integrals on paper with a pencil, but I thought it would help you to know what His Majesty Maple the Sixth says about their value:

> int(1/(a^2+x^2)^2,x=-infinity..infinity);
{ Pi
{ - 1/2 ---- a < 0
{ 3
{ a
{
{ infinity a = 0
{
{ Pi
{ 1/2 ---- 0 < a
{ 3
{ a
> int(x^2/(a^2+x^2)^2,x=-infinity..infinity);
{ Pi
{ - 1/2 ---- a < 0
{ a
{
{ infinity a = 0
{
{ Pi
{ 1/2 ---- 0 < a
{ a
> int(sin(k*x)^2/x^2,x=0..infinity);
1/2 signum(k) Pi k
>