- #1

- 23

- 0

I am trying to find an analytic solution to the following:

[tex]\int_{-1}^{1}\exp(-p\sqrt{1-x^{2}}-qx)dx[/tex]

where [tex]p,q > 0.[/tex]

Does anyone have any ideas? Thanks.

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- Thread starter appelberry
- Start date

- #1

- 23

- 0

I am trying to find an analytic solution to the following:

[tex]\int_{-1}^{1}\exp(-p\sqrt{1-x^{2}}-qx)dx[/tex]

where [tex]p,q > 0.[/tex]

Does anyone have any ideas? Thanks.

- #2

- 365

- 0

Hi! Welcome to PF!

Could you possibly tell me what exp supposed to mean ?

Regards.

Could you possibly tell me what exp supposed to mean ?

Regards.

- #3

- 501

- 2

http://www.efunda.com/math/taylor_series/exponential.cfm

Wouldn't take care of the square root, But with your range (-1,1) You could probably get away with only the first few terms. This would be an approx. though.

- #4

- 365

- 0

Try, the substitution method.

[tex]t=e^{-p\sqrt{1-x^{2}}-qx}[/tex]

Regards.

[tex]t=e^{-p\sqrt{1-x^{2}}-qx}[/tex]

Regards.

- #5

- 607

- 0

I see no reason to think there is a closed-form expression for this.

- #6

- 1,851

- 7

I am trying to find an analytic solution to the following:

[tex]\int_{-1}^{1}\exp(-p\sqrt{1-x^{2}}-qx)dx[/tex]

where [tex]p,q > 0.[/tex]

Does anyone have any ideas? Thanks.

Substitute x =sin(t) and take a look at the integral representations of Bessel functions, e.g., http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html" [Broken]

Last edited by a moderator:

- #7

- 23

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Substitute x =sin(t) and take a look at the integral representations of Bessel functions, e.g., http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html" [Broken]

Thanks Count Iblis,

I think the result does involve a Bessel function. Related to this, does anyone know how the result

[tex]\int_{0}^{2\pi}\exp(x\cos\theta + y\sin\theta)d\theta = 2\pi I_{0}(\sqrt{x^2+y^2})[/tex]

is obtained? [tex]I_{0}[/tex] is the modified Bessel function of the first kind.

Last edited by a moderator:

- #8

- 1,851

- 7

Thanks Count Iblis,

I think the result does involve a Bessel function. Related to this, does anyone know how the result

[tex]\int_{0}^{2\pi}\exp(x\cos\theta + y\sin\theta)d\theta = 2\pi I_{0}(\sqrt{x^2+y^2})[/tex]

is obtained? [tex]I_{0}[/tex] is the modified Bessel function of the first kind.

Add up the sin and cos, like:

x sin(theta) + y cos(theta) = sqrt(x^2 + y^2) cos(theta + phi)

The value of phi doesn't matter, because the integral is over an entire period. You can substitute theta = u - phi and then the integral over u will be from minus phi to 2 pi -phi, but that is the same as integrating over y from zero to 2 pi. So, you get rid of the phi this way.

And then it is just a matter of using the integral definition of I_{0}

- #9

- 23

- 0

Add up the sin and cos, like:

x sin(theta) + y cos(theta) = sqrt(x^2 + y^2) cos(theta + phi)

The value of phi doesn't matter, because the integral is over an entire period. You can substitute theta = u - phi and then the integral over u will be from minus phi to 2 pi -phi, but that is the same as integrating over y from zero to 2 pi. So, you get rid of the phi this way.

And then it is just a matter of using the integral definition of I_{0}

Excellent, thanks Count Iblis!

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