Paper 1, Section I, E

Further Complex Methods | Part II, 2011

Show that the following integral is well defined:

I(a,b)=∫0∞(e−bxeiaex−1−ebxe−iaex−1)dx,0<a<∞,a≠2nπ,n∈Z,0<b<1I(a, b)=\int_{0}^{\infty}\left(\frac{e^{-b x}}{e^{i a} e^{x}-1}-\frac{e^{b x}}{e^{-i a} e^{x}-1}\right) d x, \quad 0<a<\infty, a \neq 2 n \pi, n \in \mathbb{Z}, 0<b<1

Express I(a,b)I(a, b) in terms of a combination of hypergeometric functions.

[You may assume without proof that the hypergeometric function F(a,b;c;z)F(a, b ; c ; z) can be expressed in the form

F(a,b;c;z)=Γ(c)Γ(b)Γ(c−b)∫01tb−1(1−t)c−b−1(1−tz)−adtF(a, b ; c ; z)=\frac{\Gamma(c)}{\Gamma(b) \Gamma(c-b)} \int_{0}^{1} t^{b-1}(1-t)^{c-b-1}(1-t z)^{-a} d t

for appropriate restrictions on c,b,zc, b, z. Furthermore,

Γ(z+1)=zΓ(z)⋅]\Gamma(z+1)=z \Gamma(z) \cdot]

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