On several occasions we face contour integrals of the type

$$\begin{aligned} J=\frac{1}{2\mathrm{i}}\int _{C_{u}}\frac{F(u)}{\sqrt{\cosh s-\cosh u}}\, \mathrm{d}u \qquad s>0 \, ,\end{aligned}$$

(3.131)

where the path of integration

\(C_{u}\) in the complex plane

*u* is defined like in (

2.58), recall, it consists of three sides of the infinite rectangle with vertices at

\(\infty -\pi \mathrm{i},\,-\pi \mathrm{i},\,\pi \mathrm{i},\,\infty +\pi \mathrm{i}\), run in this order. It is supposed that function

*F*(

*u*) is analytic in the half-strip

\(\Pi \), surrounded by the contour

\(C_{u}\), and is decreasing fast at

\(\mathfrak {R}u\rightarrow \infty \) (typically

\(F(u)\propto \mathrm{e}^{-\frac{u^{2}}{2t}}\)). Function

$$\begin{aligned} f(u)=\left( \cosh s - \cosh u\right) ^{1/2} \, , \end{aligned}$$

(3.132)

defined originally on contour

\(C_{u}\), can be continued analytically into the half-strip

\(\Pi \), equipped with the cut along the real axis from the branching point

\(u=s\) to

\(+\infty \). The cut allows to select a single valued analytic branch of

*f*(

*u*), fixed by the condition of

*f*(

*u*) being positive on

\(C_{u}\), as discussed in Sect.

3.2.

According to the Cauchy theorem, the value of integral *J* of the analytic function *F*(*u*) / *f*(*u*) depends only on the ends of the path \(C_{u}\). Even more, the ends themselves may be shifted along the vertical line \(\mathfrak {R}u=\infty \) due to decreasing *F*(*u*). This means that without changing the value of integral *J* we may choose any path *C* that sustains the global properties of the original contour \(C_{u}\); it starts at \(\mathfrak {R}u=+\infty \) on or below the lower bank of the cut, terminates at \(\mathfrak {R}u=+\infty \) on or above the upper bank, and goes around the cut in the negative direction (crossing cut is not allowed). This said, we squeeze contour \(C_{u}\), placing integration directly onto the banks of the cut, plus an infinitesimal circumference surrounding point \(u=s\). Contribution from the latter vanishes as its radius tends to zero, because the singularity \(1/f(u) \propto (\delta u)^{-1/2}\) is integrable. This leaves only two integrals, taken in opposite directions along the banks.

We need now values of function

*f*(

*u*) on the cut banks. On either bank,

\(u= \mathfrak {R}u>s\) so that

*f*(

*u*) (

3.132) is pure imaginary. Proper signs can be determined by moving from contour

\(C_{u}\) to the cut banks. Take

*u* on the far right end of the upper leg of

\(C_{u}\),

\(u=R+\mathrm{i}\pi , \quad R\gg s\), where asymptotics of

*f*(

*u*) (

3.132) is

\(\sqrt{2} f(u) \approx \mathrm{e}^{R/2} = -\mathrm{i}\mathrm{e}^{u/2}\). By continuity, this asymptotics remains valid, when moving down off

\(C_{u}\) and reaching the upper bank at

\(u=R\) with

\(\sqrt{2} f(u) \approx -\mathrm{i}\mathrm{e}^{R/2} \). We conclude that on the upper bank

$$ f(u)=-\mathrm{i}\left| f(u)\right| =-\mathrm{i}\sqrt{\cosh u -\cosh s} $$

In the same manner, moving upward from the far right end of the lower leg of

\(C_{u}\), we find that on the lower bank

$$ f(u)=+\mathrm{i}\sqrt{\cosh u -\cosh s} $$

(We write

\(\sqrt{(\cdot )}\), rather than

\((\cdot )^{1/2}\), to underline that the expression under square root is positive and that the positive value of the square root is taken).

Signs of

*f*(

*u*) and directions of integration are opposite on the two banks, meaning that contributions from the two are equal, and the whole integral

*J* is twice that along the upper bank,

$$\begin{aligned} J=\frac{1}{2\mathrm{i}}\int _{C_{u}}\frac{F(u)}{\sqrt{\cosh s-\cosh u}}\, \mathrm{d}u =\int _{s}^{\infty }\frac{F(u)}{\sqrt{\cosh u-\cosh s}}\, \mathrm{d}u \end{aligned}$$

(3.133)

We use this transformation when computing PDFs of normal, log-normal, and general SABR with zero correlation; the moment generating function of the inverse random time

\(\tau \); and option values. In the case of Log-normal SABR the integral to be transformed is slightly different from (

3.131), but the consideration follows the same lines. The original and transformed integrals look like follows

$$\begin{aligned} \tilde{J}&=\frac{1}{2\mathrm{i}}\int _{C_{u}}\frac{F(u)\, E^{-\lambda \sqrt{\cosh s-\cosh u}}}{\sqrt{\cosh s-\cosh u}} \, \mathrm{d}u \nonumber \\&= \int _{s}^{\infty }\frac{F(u)\cos (\lambda \sqrt{\cosh u-\cosh s})}{\sqrt{\cosh u-\cosh s}} \, \mathrm{d}u \, .\end{aligned}$$

(3.134)