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Bessel equation

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If m ∈ Z, then I−m(z) and Im(z) are linearly independent and span the space of solutions of the<br />

modified <strong>Bessel</strong> <strong>equation</strong>.<br />

We have<br />

Im(e ±iπ z) = e ±iπm Im(z).<br />

1.4 Integral representations of modified <strong>Bessel</strong> function<br />

We start with <strong>Bessel</strong>-Schläfli-type representations. If Rez > 0, then the basic contour integral representation<br />

of Im is<br />

Im(z) = 1<br />

<br />

2πi ]−∞,0 + <br />

z<br />

exp<br />

,−∞[ 2 (t + t−1 <br />

) t −m−1 =<br />

dt (1.4.5)<br />

1<br />

<br />

z<br />

<br />

<br />

m<br />

exp s +<br />

2πi 2<br />

z2<br />

<br />

s<br />

4s<br />

−m−1 ds. (1.4.6)<br />

]−∞,0 + ,−∞[<br />

To see this note that by (1.4.5), Im is holomorphic around zero. Besides, by the Hankel identity<br />

<br />

1 1<br />

= exp (s) s<br />

Γ(m + 1) 2πi<br />

−m−1 ds.<br />

We also have<br />

]−∞,0 + ,−∞[<br />

I−m(z) = 1<br />

<br />

2πi [(0−0) + <br />

z<br />

exp<br />

] 2 (t + t−1 <br />

) t −m−1 dt. (1.4.7)<br />

Here, the contour starts at 0 from the negative side on the lower sheet, encircling 0 in the positive<br />

direction and ends at 0 from the negative side on the upper sheet,<br />

To see this we make a substitution t = s −1 in (1.4.5) noting that dt = −s −2 ds, and then we change<br />

the orientation of the contour.

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