Using complex exponentials

[aks_sky]

∫_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 don't know if it makes any difference if the cos is replaced by a sin.. does it make any difference?

cheers

 

[Dick]

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.

 

[aks_sky]

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)

 

[Cyosis]

Are you claiming that $\cos ax+i \sin ax=\cos a x$? This is of course not true. Try $\sin x=Im(e^{i x})$.

 

[HallsofIvy]

yup, because if i use cos (ax) i can replace it with e ^ i(ax) because i can use the complex exponential there.

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.

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)

 

[aks_sky]

ahhh i get it.. sweet.. thank u