
#1
May609, 11:04 PM

P: 56

∫_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 



#2
May609, 11:12 PM

Sci Advisor
HW Helper
Thanks
P: 25,168

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.




#3
May609, 11:22 PM

P: 56

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^∞ ex (cos ax + isin ax) dx))) Therefore we get: ∫_0^∞ e^(1ia)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) 



#4
May709, 02:09 AM

HW Helper
P: 1,495

using complex exponentials
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].




#5
May709, 06:06 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,881





#6
May709, 06:23 AM

P: 56

ahhh i get it.. sweet.. thank u



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