Question

evaluate the integral (haven't done this in awhile so please help if you can)

Answer 1

\[I=\int _{0}^{\infty }e^{\alpha x}sinxdx\]
\[(\alpha>0)\]

Answer 2

\[I=lim_{b\rightarrow \infty}\int _{0}^{b }e^{\alpha x}sinxdx\]
right?

Answer 3

yup

Answer 4

now i don't know what to do next

Answer 5

for the integral only
just do parts. let u = sinx, dv=e^(alpha x)dx

Answer 6

du = cos x, right? \(v= \dfrac{e^{\alpha x}}{\alpha}\)

Answer 7

dx*

Answer 8

:) thanks @Nnesha

Answer 9

\[u=sinx, dv=e^{\alpha x}dx\]
\[du=cosxdx, v=\frac{e^{\alpha x}}{\alpha}\]
\[I=lim_{b\rightarrow \infty}sinx*\frac{e^{\alpha x}}{\alpha}|^{b}_{0}-(\int^{b}_{0} \frac{e^{\alpha x}}{\alpha}cosxdx)\]

Answer 10

isn't that supposed to be u=e^axhmmm?

Answer 11

good

Answer 12

do it again with the integration, what do you get?

Answer 13

oh, zepdrix is here. He is an expert of integration. Let's wait for him.

Answer 14

Are you supposed to solve this integral using complex numbers? :)
Just asking cause of the class you're taking :3 hehe

Answer 15

yeah considering the other course work I have i probably do have to use complex numbers. havent went over this yet so i don't know what to do

Answer 16

Ahhh I gotta go right now D:
I can explain it later if these rascals don't beat me to it.

Answer 17

Here!! since zepdrix went. I give you my work but you have to finish the problem by yourself. Just take lim when b goes to infinitive.

Answer 18

@zepdrix if you have a way using complex numbers i'll wait to see it, but i only have till 9:00 pm MST

Answer 19

@Loser66 thank you what you did looks to make a lot of sense. I'm still working it out for myself just for my sake

Answer 20

don't you want \(\alpha<0\)

Answer 21

\[\Large\sin x = {e^{ix} - e^{-ix} \over 2i}\]

Answer 22

So uhhh.. ya... here is a way to do it :x
I kinda ramble on for a while...
but just in case you wanted to see the steps, they're included.

So like.. we know this,\[\large\rm e^{ix}=\cos x+i \sin x\]So the real portion of this complex number is only the cosine, ya?\[\large\rm \color{orangered}{Re(e^{ix})=\cos x}\]While the imaginary part is the sine function,\[\large\rm \color{royalblue}{Im(e^{ix})=\sin x}\]So for your integral,\[\large\rm I=\int\limits e^{\alpha x}\color{royalblue}{\sin x}~dx\]We can say that this is instead the imaginary part of our complex exponential,\[\large\rm I=\int\limits\limits e^{\alpha x}\color{royalblue}{Im(e^{ix})}~dx\]So that allows us to combine our exponentials,
\[\large\rm I=Im\int\limits e^{\alpha x}e^{ix}dx\]Into one exponential,\[\large\rm I=Im\int\limits\limits e^{\alpha x+ix}dx\]And then pick out the imaginary parts in the end for our solution.\[\large\rm I=Im\int\limits\limits e^{(\alpha+i)x}dx\]
So you would just integrate your exponential,\[\large\rm I=Im\frac{1}{\alpha+i}e^{(\alpha+i)x}\]Multiply by conjugate,\[\large\rm I=Im\frac{1}{\alpha+i}\left(\frac{\alpha-i}{\alpha-i}\right)e^{(\alpha+i)x}\]giving us,\[\large\rm I=Im\frac{\alpha-i}{\alpha^2+1}e^{(\alpha+i)x}\]Breaking up our exponentials again,\[\large\rm I=Im\frac{\alpha-i}{\alpha^2+1}e^{\alpha x}e^{ix}\]expanding out our complex exponential in terms of cosine and sine,\[\large\rm I=Im\frac{\alpha-i}{\alpha^2+1}e^{\alpha x}\left(\cos x+i \sin x\right)\]Ok and so like.. we only want the imaginary parts,
Which is going to be the alpha/(stuff) times the isinx,
and also the -i/(stuff) times the cosx.\[\large\rm I=Im~e^{\alpha x}\left[\frac{\alpha }{\alpha^2+1}i \sin x-\frac{i}{\alpha^2+1}\cos x+(real~stuff)\right]\]So the imaginary stuff is these two terms,\[\large\rm I=e^{\alpha x}\left[\frac{\alpha }{\alpha^2+1}\sin x-\frac{1}{\alpha^2+1}\cos x\right]\]

Answer 23

oh ok that actually makes a lot of sense. I'm a bit fuzzy on how exactly you got the last part from the second to last part and do i still evaluate the integral with the lim as b approaches infinity?

Answer 24

also \[e^{-\alpha x}\]

Answer 25

i forgot the negative *facepalm*

Answer 26

Oh noesss you're out of time XD

Answer 27

just about D=

Answer 28

\[\large\rm I=e^{-\alpha x}\left[\frac{-\alpha }{\alpha^2+1}\sin x-\frac{1}{\alpha^2+1}\cos x\right]\]Both terms are negative, pull out the negative,\[\large\rm I=-e^{-\alpha x}\left[\frac{\alpha }{\alpha^2+1}\sin x+\frac{1}{\alpha^2+1}\cos x\right]\]And then ya evaluate at the limits I guess.

Answer 29

You can always use Wolfram to ... *cough* CHECK YOUR WORK if you're in a pinch ;O

Answer 30

got it thanks

Answer 31

@zepdrix thank you!