Good mourning, can anyone help me whith this kind of improper integral, there is the so called the Cauchy's singularity that is similar to the above integral but with (t-x) that give 1/Lambda*(t-x). thanks
Dear Brick A.A.,
should not be -y in the exponential part?
The integrating by parts gives a primitive function in the case of a positive exponent as well, but I think the integral does not exist (oscilations).
For the exponent (-y*lambda), I obtained the result 1/(t-x) if the sine has (t-x) in argument, and 1/(t+x) for the argument with (t+x).
I do not unterstand your last words "that give 1/Lambda*(t-x)"
Thanks Mr Stoica and Miss Cernanova for your answers,there is an error in the above integral and i have the -y (as you have said Miss Cernanova), the integral above with exponential of -Lambda*y give the results you have mentioned Miss in your answer for (t+x) and for (t-x), thanks a lot
Dear Brick,
instead of integration in part I would draw your attention that
\[ I(y) = \int_0^\infty e^{-\lambda y} e^{i (t+x) \lambda} d\lambda = \frac1{y-i(t+x)}; \]
Now it is enough to consider the imaginary part of $\lim_{y \to 0+}I(y)$ which gives the result $(t+x)^{-1}$ quoted by Dr Cernanova.
Best,
Tibor Pogany
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