Good mourning, i used the demostration shown in the figure below to reduce the integral shown to a simple form, the function Psi is an odd function with respect to x independant variable. I just want to check whether it is right or not, could any one help?
@ Brick
Possible Error in 3rd line, first term in the RHS.
What kind of operation is that? I guess, you replaced t by -t.
Then limits -a and 0 become a and 0 which is correct. Denominator t - x becomes - t - x, again correct. But Psi is an odd function as claimed by you, so that psi (-t) = - psi (t).
There should be a negative sign before the first term in the RHS of 3rd line. Correct it and subsequent equations accordingly.
Thanks Mr Pandey for your help, i have used the oddiness of the function psi but didn't show that on the picture, the sign (-) that came from psi cancels with that came from (-t-x).
Brick
Sorry, my mistake. You are right, Another minus comes from dt part too!
seems that i didn't catch all minus fishes :DDD, thanks Mr Pandey
Thank you Mr Raul, i missed something, when integrating from 0 to a it is similar to integrating from a to 0 but we should multiply by (-1). this make it correct
When doing it, I took that into account, too, and the whole thing works.
Does x lie between -a and a? Then there could be a difficulty, because you have zero in the denominator.
Unfortunately the integrals are all undefined due to the singularity of the integrand for t = x if x belongs to the interval [-a,a] (I assume that this is the case of interest. Johann's correct remark was not yet there when I first wrote my answer).
As a result, an application of the normal calculus rules is not guaranteed to transform equals to equals. For instance for the special case psi(t) = t (it is in most cases advantageous to look for simple special cases!) Mathematica gives for your original integral an ArcTanh-expression and for your simplified version the result 'undefined'.
For singular Integrals of your kind the textbooks have a chapter on Cauchy's principal value ...
You can learn also that not all green lights from experts can be trusted.
You are right, this integral is a singular part of Cauchy type that somtimes apear in fracture mechanics when we use the integral equation method and this part can be solved using the Cauchy's principal value, the integral equation is mostly singular and need a special numerical schemas to be solved, in my above question i don't talk about the integral itself but about the simplification i made to reduce it to that simple part that is of Cauchy type, integrating it is a part of solving the SIE that involve it. thanks
OK, I omitted consideration of the location of the constant x. (I guess I asssumed it was outside the interval [- a,a].) But then, I am no expert...
One more comment: since t is an odd function, just as psi(t) is, their product is an even function; on tthe other hand, the denominator , tsquared - xsquared, is also even. Therefore, the integral sought is double the integral of the same integrand in the interval [-a, 0] or [0,a]. (Since the integral is already solved, I don't think this can be of any use, though.)
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