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skyflyer54
Sep23-06, 11:23 PM
Hi, having difficulty solving the following question.. Any1 can help me and guide me through solving it ?? or is there any good website or reference to recommand ?? Thankz alot..

Data
Sep24-06, 12:20 AM
The integral is undefined. However, if you were to define it, there's a particular value that seems pretty natural. To help you guess what that might be, notice that x(t) is an odd function:

x(-t) = (-t)^3\cos (-10t)^3 = -t^3\cos (10t)^3 = -x(t).

Does that help? :smile:

(the "natural" value for the integral that I'm speaking of is the Cauchy principal value)

skyflyer54
Sep24-06, 12:38 AM
Oh.. I only know the shortcut method but don't know whether u are refering to it or not...

I only know that 't' is odd function and cos(10t) is even function.. so multiply odd with even will give me odd function...

I'm confused is the part on how to solve it mathematically said integration by parts.. haha ...

d_leet
Sep24-06, 12:44 AM
Oh.. I only know the shortcut method but don't know whether u are refering to it or not...

I only know that 't' is odd function and cos(10t) is even function.. so multiply odd with even will give me odd function...

I'm confused is the part on how to solve it mathematically said integration by parts.. haha ...

To integrate that function integration by parts would work, but would probably be extremely messy, the problem you will run into is evaluating the integral since both limits will be infinite and the improper integral will be undefined as Data said above.

jpr0
Sep24-06, 03:25 AM
If you wanted to use integrate by parts you could probably set the upper and lower limits of integration to some constant \pm a. This symmetric choice of limits will allow you to evaluate the integral by parts, by utilising the symmerty properties of the the now-integrated function to determine what cancellations occur etc. Then in the final step take the limit of |a|\to\infty.

murshid_islam
Sep24-06, 06:42 AM
what about contour integration?

Data
Sep24-06, 09:22 AM
If you wanted to use integrate by parts you could probably set the upper and lower limits of integration to some constant \pm a. This symmetric choice of limits will allow you to evaluate the integral by parts, by utilising the symmerty properties of the the now-integrated function to determine what cancellations occur etc. Then in the final step take the limit of |a|\to\infty.

That is precisely taking the Cauchy principal value. It is trivial to see what the result will be, because x is odd.

skyflyer54
Sep24-06, 08:31 PM
Okie.. Thank every1 for the guide and comment given.

murshid_islam
Sep25-06, 08:59 AM
shouldn't the result be zero since the function is odd?

HallsofIvy
Sep25-06, 09:05 AM
Yes, for any A,
\int_{-A}^{A} t^3 cos^3(10t)dt= 0
so the limit as A->infinity, the "Cauchy Principal Value" would be 0. That's what everyone has been saying.

Data said that integral is undefined because the actual improper integral must be
\int_{-A}^B x^3 cos^3(10t)dt
with A and B going to infinity independently. That is undefined.

skyflyer54
Sep25-06, 05:12 PM
tat true.. if the limit is from A to -A or even from Infinity to -Infinity, the result will be zero. If it is undefined, say like limit from A to B den the answer should not be zero. right ?

This question in the first place was just to solve it to be odd function using mathematical approach. kekez..

Did I said anything wrong ??? If yes, pls correct mi. thankz.. heez.

HallsofIvy
Sep25-06, 05:49 PM
tat true.. if the limit is from A to -A or even from Infinity to -Infinity, the result will be zero. If it is undefined, say like limit from A to B den the answer should not be zero. right ?

This question in the first place was just to solve it to be odd function using mathematical approach. kekez..

Did I said anything wrong ??? If yes, pls correct mi. thankz.. heez.
Yes, the "even from infinity to -infinity" is wrong. As I said before
\int_{-\inty}^\infty f(x) dx= \lim_{A\rightarrow -\infty}\lim_{B\rightarrow \infty} \int_A^B f(x)dx

Certainly, if an integral is "undefined", then it is not 0! The original question was to find the value of that integral if it existed. I don't know what you mean by "solve it to be odd function".

skyflyer54
Sep26-06, 07:36 AM
From observation, we already know that the function is an odd function, but What i mean is solve and proof it that the equation is an odd function. what i mean is solve and proof it mathematically if possible.

HallsofIvy
Sep26-06, 07:58 AM
From observation, we already know that the function is an odd function, but What i mean is solve and proof it that the equation is an odd function. what i mean is solve and proof it mathematically if possible.
I have absolutely no idea what you mean. You say "What i mean is solve and proof it that the equation is an odd function" but you had just said "we already know that the function is an odd function". Prove what mathematically?
That the integral given doesn't exist?

That the Cauchy Principal Value is 0?

skyflyer54
Sep26-06, 08:44 PM
I have absolutely no idea what you mean. You say "What i mean is solve and proof it that the equation is an odd function" but you had just said "we already know that the function is an odd function". Prove what mathematically?
That the integral given doesn't exist?

That the Cauchy Principal Value is 0?

ok.. thank.. maybe i should read up cachy principal again..