Analytic solution to an exponential integral

[appelberry]

Hello,

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

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

where p,q > 0.

Does anyone have any ideas? Thanks.

 

[Дьявол]

Hi! Welcome to PF!

Could you possibly tell me what exp supposed to mean ?

Regards.

 

[flatmaster]

Taylor series expansion of the exponential.
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.

 

[Дьявол]

Try, the substitution method.

t=e^{-p\sqrt{1-x^{2}}-qx}

Regards.

 

[g_edgar]

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

 

[Count Iblis]

Hello,

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

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

where p,q > 0.

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"

 

[appelberry]

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

Thanks Count Iblis,

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

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

is obtained? I_{0} is the modified Bessel function of the first kind.

 

[Count Iblis]

Thanks Count Iblis,

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

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

is obtained? I_{0} 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}

 

[appelberry]

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!