632 CHAPTER 32. SOLUTIONS NEAR A REGULAR SINGULAR POINT

Proof: It remains to obtain the above addition formula. This is remarkably easy toobtain.

e((x+y)/2)(t−1/t) =∞

∑m=−∞

tmJm (x+ y)

e((x+y)/2)(t−1/t) = e(x/2)(t−1/t)e(y/2)(t−1/t)

=∞

∑l=−∞

t lJl (x)∞

∑k=−∞

tkJk (y)

and in this product, the tm term is the sum of products for which l + k = m. That is,

Jm (x+ y) =∞

∑k=−∞

Jk (y)Jm−k (x)

This shows the addition formula. ■Of course t was completely arbitrary as long as it is not zero. Thus let it equal eiθ in

32.23. Then from Euler’s identity, eiθ = (cos(θ)+ isin(θ)) ,

e(x/2)(2isinθ) =∞

∑m=−∞

(eiθ)m

Jm (x)

Then using Euler’s identity again,

cos(xsin(θ))+ isin(xsin(θ)) =∞

∑m=−∞

(cos(mθ)+ isin(mθ))Jm (x)

Equating real and imaginary parts,

cos(xsin(θ)) =∞

∑m=−∞

cos(mθ)Jm (x)

sin(xsin(θ)) =∞

∑m=−∞

sin(mθ)Jm (x)

Now recall from trig. identities,

cos(a)cos(b)+ sin(a)sin(b) = cos(a−b)

multiply the top by cos(nθ) and the bottom by sin(nθ) and add. Thus

cos(nθ − xsin(θ)) =∞

∑m=−∞

cos(nθ −mθ)Jm (x)

Because of the uniform convergence of the partial sums of the above series which followsfrom computations like those in 32.22, one can interchange

∫π

0 with the infinite summation.This yields ∫

π

0cos(nθ − xsin(θ))dθ = πJn (x)

because, unless n = m,∫

π

0 cos(nθ −mθ)dθ = 0. Therefore, this yields the very importantintegral identity for Jn (x) ,

Jn (x) =1π

∫π

0cos(nθ − xsin(θ))dθ (32.25)