∫_0^∞ e^-x sin (ax) dx integrating from 0 to infinity for e^-x sin (ax) i was wondering that if i use complex exponentials for this will it be the same as solving for ∫_0^∞ e^-x cos (ax) dx. Will it make any difference if there is sin instead of cos because i know how to solve for cos (ax) but i dont know if it makes any difference if the cos is replaced by a sin.. does it make any difference? cheers
It makes a difference in the sense that the answer is different, yes. The technique is pretty much the same, if that's what you mean.
yup, because if i use cos (ax) i can replace it with e ^ i(ax) because i can use the complex exponential there. But what do i use for sin (ax).. i will just do what i did for cos (ax) ∫_0^∞ e^-x. e^i(ax) ..... (((since ∫_0^∞ e-x (cos ax + isin ax) dx))) Therefore we get: ∫_0^∞ e^-(1-ia)x dx then i can go ahead from there to get 1/ 1 + a^2.... now the only thing i am confused with is that this is for the complex exp. for cos (ax) what will it be for sin (ax)
Are you claiming that [itex]\cos ax+i \sin ax=\cos a x[/itex]? This is of course not true. Try [itex]\sin x=Im(e^{i x})[/itex].
You can't just "replace it with e^i(ax)". e^(iax)= cos(x)+ i sin(x). You can replace cos(x) with e^i(ax) and then just take the real part of the answer. And, of course, with sin(ax) you can replace it with e^i(ax) and take only the imaginary part of the answer.