I am struggling to find the integration of the expression below,∫−π2π2cos⁡(acos⁡θ)eimθe−ibsin⁡θdθwhere a and b are arbitrary constant and m is an integer.

You integral is 12(I(a,b)+I(−a,b)) whereI(a,b)=∫−π2π2exp⁡(imθ)exp⁡(iacos⁡θ−ibsin⁡θ),dθand by writing acos⁡θ−bsin⁡θ as ρcos⁡(θ+φ), with ρ=a2+b2 and φ=arctan⁡baI(a,b)=e−imφ∫φ−π2φ+π2exp⁡(imθ)exp⁡(iρcos⁡(θ)),dθwhereeiρcos⁡θ=J0(ρ)+2∑n≥1inJn(ρ)cos⁡(nθ)gives:I(a,b)=e−imφ[π,J0(ρ)+2∑n≥1inJn(ρ)∫φ−π2φ+π2exp⁡(imθ)cos⁡(nθ),dθ]and your integral is represented as a fast-convergent series.

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Jaylin Clements

Answered question

2022-04-01

I am struggling to find the integration of the expression below,

\int_{- \frac{π}{2}}^{\frac{π}{2}} \cos (a \cos θ) e^{i m θ} e^{- i b \sin θ} d θ

where a and b are arbitrary constant and m is an integer.

Answer & Explanation

alwadau8ndv

Beginner2022-04-02Added 9 answers

You integral is

\frac{1}{2} (I (a, b) + I (- a, b))

where

I (a, b) = \int_{- \frac{π}{2}}^{\frac{π}{2}} \exp (i m θ) \exp (i a \cos θ - i b \sin θ), d θ

and by writing

a \cos θ - b \sin θ

ρ \cos (θ + φ)

, with

ρ = \sqrt{a^{2} + b^{2}}

and

φ = \arctan \frac{b}{a}

I (a, b) = e^{- i m φ} \int_{φ - \frac{π}{2}}^{φ + \frac{π}{2}} \exp (i m θ) \exp (i ρ \cos (θ)), d θ

where

e^{i ρ \cos θ} = J_{0} (ρ) + 2 \sum_{n \geq 1} i^{n} J_{n} (ρ) \cos (n θ)

gives:

I (a, b) = e^{- i m φ} [π, J_{0} (ρ) + 2 \sum_{n \geq 1} i^{n} J_{n} (ρ) \int_{φ - \frac{π}{2}}^{φ + \frac{π}{2}} \exp (i m θ) \cos (n θ), d θ]

and your integral is represented as a fast-convergent series.

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