Integration problem, completely stuck

I've been stuck on this question for the last 40 minutes and while it seems fairly straightforward I just can't get anywhere with it and so thought I would ask the wonderful people of TSR for some help!

Also, please excuse me typing it out, I never managed to figure out how to use LATEX.

I need to find the integral from 0 to infinity of exp(-y)cos(xy)dy

As far as i'm aware using integration by parts is a must here so I took the following:

u = exp(-y)
v' = cos(xy)
SO
u' = -exp(-y)
v = [sin(xy)]/x

which gives (using I to stand for the integral from 0 to infinity)

[exp(-y)sin(xy)]/x - I[-exp(-y)sin(xy)]/x

From then I need to use parts again and I take:

u = sin(xy)/x
v' = -exp(-y)/x
SO
u' = cos(xy)
v = exp(-y)/x

Which then gives me:

[exp(-y)sin(xy)]/x - [[exp(-y)sin(xy)]/x^2 - I[cos(xy)exp(-y)/x]]

But from here it all seems to fall apart.

I know what the answer needs to be, I know it has to be:

exp(-y)[xsin(xy)-cos(xy)]
------------------------------
1+x^2

but I have no idea how to get to there.

Thanks very much.

Reply 1

$Avatar for Get me off the £\?%!^@ computer$

Get me off the £\?%!^@ computer

5

Original post by mackemforever

I never managed to figure out how to use LATEX.

http://www.thestudentroom.co.uk/wiki/LaTex

Give it five minutes.

I expect you already know, but after doing integration by parts, twice, the original integral will reappear.

Call it I and you get

I=stuff -x^2 I

Reply 2

Pheylan

Original post by mackemforever

I need to find the integral from 0 to infinity of exp(-y)cos(xy)dy

seems quite impossible (without doing something fancy that i've not learned yet)

Reply 3

ak9779

9

When you do by parts the second time, you will end up with your original integral. If you say your initial integral = I then you get:

I = whatever - I

therefore 2I = whatever

therefore I = whatever/2

Reply 4

mackemforever

OP

17

Original post by ak9779

When you do by parts the second time, you will end up with your original integral. If you say your initial integral = I then you get:

I = whatever - I

therefore 2I = whatever

therefore I = whatever/2

The problem being that I kept on ending up with I = whatever + I, therefore the left side vanishes.

That was why I was asking for somebody to run through it.

Reply 5

Oh I Really Don't Care

17

feynman integration

Reply 6

mph10

i can't understand what the question is, does the exp mean

10^{-y}

??

Reply 7

mackemforever

OP

17

Original post by mph10

i can't understand what the question is, does the exp mean

10^{-y}

??

no, it means the exponential of -y, e.g. e^(-y) if that makes more sense.

Reply 8

mph10

Original post by mackemforever

no, it means the exponential of -y, e.g. e^(-y) if that makes more sense.

so the original post is

cos(xy)^{-y}

or is it

e^{-y}cos(xy)

?

Reply 9

boromir9111

17

Original post by DeanK22

feynman integration

Well, that's something I've never heard of lol....

Reply 10

mackemforever

OP

17

Original post by mph10

so the original post is

cos(xy)^{-y}

or is it

e^{-y}cos(xy)

?

e^{-y}cos(xy)

Reply 11

marcusmerehay

17

Are you integrating w.r.t. y?

EDIT: I've got the answer you've given above.

The last line before the answer should read something like:

I = [e^{-y}xsin(xy)-e^{-y}cos(xy)]-x^2I

I strongly advise checking your signs in the steps in the middle.

(edited 13 years ago)

Reply 12

SomethingWitty

Original post by mackemforever

are x and y independent variables?

Reply 13

mph10

i got to the same step as you, i think it just keeps on repeating integrals, so there must be another method, which unfortunately i cant remember

Reply 14

Farhan.Hanif93

18

Original post by mackemforever

I've been stuck on this question for the last 40 minutes and while it seems fairly straightforward I just can't get anywhere with it and so thought I would ask the wonderful people of TSR for some help!

Also, please excuse me typing it out, I never managed to figure out how to use LATEX.

I need to find the integral from 0 to infinity of exp(-y)cos(xy)dy

As far as i'm aware using integration by parts is a must here so I took the following:

u = exp(-y)
v' = cos(xy)
SO
u' = -exp(-y)
v = [sin(xy)]/x

which gives (using I to stand for the integral from 0 to infinity)

[exp(-y)sin(xy)]/x - I[-exp(-y)sin(xy)]/x

From then I need to use parts again and I take:

u = sin(xy)/x
v' = -exp(-y)/x
SO
u' = cos(xy)
v = exp(-y)/x

Which then gives me:

[exp(-y)sin(xy)]/x - [[exp(-y)sin(xy)]/x^2 - I[cos(xy)exp(-y)/x]]

But from here it all seems to fall apart.

I know what the answer needs to be, I know it has to be:

exp(-y)[xsin(xy)-cos(xy)]
------------------------------
1+x^2

but I have no idea how to get to there.

Thanks very much.

Is this part of a double integral? If so we can treat x as if though it's a constant.
You should have let u be the exponential function on both occasions when you carried out the integration by parts. By swapping over for the second integration by parts, you are essentially reverting to what you started with. Also after dealing with the limits, your results should not involve y so your above answer that you seem to have been 'given' is completely wrong. If you do as I've said above, you should get to:

\displaystyle\int^{\infty}_0 e^{-y}\cos (xy)dy = \frac{1}{x}\left[e^{-y}\sin (xy)\right]^{\infty}_0 - \frac{1}{x^2}\left[e^{-y}\cos (xy)\right]^{\infty}_0 - \frac{1}{x^2}\displaystyle\int^{\infty}_0e^{-y}\cos (xy)dy

.
Which is almost what you got but not quite. If you let:

I=\displaystyle\int^{\infty}_0 e^{-y}\cos (xy)dy

And rearrange the result I've written above for I, you have successfully carried out the integration (as soon as you evaluate the limits that result)

(edited 13 years ago)

Reply 15

Farhan.Hanif93

18

Original post by marcusmerehay

I strongly advise checking your signs in the steps in the middle.

This too, the signs in this probably get very confusing. Particularly when you're without paper... :colonhash:

Reply 16

whiplash

10

Original post by mackemforever

I
u = exp(-y)
v' = cos(xy)
SO
u' = -exp(-y)
v = [sin(xy)]/x

Something they really should teach in A level:

http://en.wikipedia.org/wiki/Integration_by_parts#Liate_rule

the exponential is always v' so just switch your u and v' assignment and it should solve your infinite recursion problem.

Reply 17

mph10

above post by Farhan.Hanif93 is correct^^^^

(edited 13 years ago)

Reply 18

mph10

Original post by whiplash

Something they really should teach in A level:

http://en.wikipedia.org/wiki/Integration_by_parts#Liate_rule

the exponential is always v' so just switch your u and v' assignment and it should solve your infinite recursion problem.

that does not work in this situation, try it, you will see.

Reply 19

Farhan.Hanif93

18

Original post by whiplash

Something they really should teach in A level:

http://en.wikipedia.org/wiki/Integration_by_parts#Liate_rule

the exponential is always v' so just switch your u and v' assignment and it should solve your infinite recursion problem.

That's not true and I've had this discussion with someone else on here before. It doesn't matter whether you choose the trig function or the exp function as u, as long as you choose the same type of function for u when you integrate by parts for a second time.

Integration problem, completely stuck

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