# Integration of exponential of inverse tangent

1. Dec 14, 2011

### Elsasw

Hi all, i need help with integration of exponential of inverse tangent, could not find it in table of integrals

the whole equation is
I=∫A/[w(a+z^2)^1/2]*exp(zb)*exp[(-i*arctan(z/(a)^1/2)]

-am trying to integrate by parts but stuck at the arctan part

Thank you.

2. Dec 14, 2011

### Mute

You won't be able to find a closed form in terms of elementary functions. It looks like the answer will necessarily involve the special function $\mbox{Ei}(z)$, defined for real numbers as

$$\mbox{Ei}(x) = \int_{-\infty}^x \frac{e^t}{t},$$
though your integral will require the extension of this function on the complex plane.

Anyways, for your problem, I would use $e^{-i\theta} = \cos\theta - i \sin\theta$. Since $\theta = \tan^{-1}(z/\sqrt{a})$ you can rewrite the sine and cosine in terms of only z, a, and (z^2+a)^(0.5). The last hint I will give is that the overall denominator you get can be factored (with imaginary roots) to cancel with part of the numerator.

3. Dec 15, 2011

### Elsasw

Thank you mute. I redid the initial equations and put the inverse tangent term as arctan (z/a) instead of arctan (z/a^1/2) as earlier.ignore the w as i already factor it out. so now i got

I=∫A/[(z^2+a^2)^1/2]*exp(bz)exp(-i*arctan(z/a)) *dz

=A∫1/[(z^2+a^2)^1/2]*exp(bz)*cos(theta)-i*sin(theta) *dz

where theta=arctan(z/a)

=A∫1/[(z^2+a^2)^1/2]*exp(bz)*(a-i*z)/[(z^2+a^2)^1/2]*dz

finally

I=A∫(1/a+iz)exp(bz)*dz

After this, let u=a+iz

I=-iA*exp(iab)*∫exp(-ibu)/u*du

This is referring the exponential integral term rite. I hope this is what you meant.
Thanks again.

4. Dec 15, 2011

### Mute

I haven't checked it carefully, but yes, that looks about right. You can now identify

$$\int du~\frac{e^{-ibu}}{u} = \mbox{Ei}(-ibu) + \mbox{const}$$
and put u back in terms of z.

5. Dec 15, 2011

### Elsasw

Thanks alot mute...