Contour of integration

Now let us introduce one useful trick, so-called closed time-path contour of integration. First, note that the expression of the type... 

A being a contour of integration encircling t and t" but no other transition point, with the integration performed in the direction that makes Kq positive when t and t" are real, i.e., when the barrier is superdense, but negative when t and t" are complex conjugate, i.e., when the barrier is underdense. 

Fig. 5.1a. Qualitative behavior of —Q2( ) for m / 0. The wavy line indicates a cut, and is a closed contour of integration, on which the phase of Q1,/2( ) is indicated. The point 0 lies to the left of the origin.

Fig. 5.1b. Qualitative behavior of —Q2(rf) for m / 0 in the sub-barrier case. The wavy lines are cuts, and Al and Ak are closed contours of integration. The part of Al that lies on the second Riemann sheet is drawn as a broken line. The phases of Q1 2(r ) indicated in the figure refer to the first Riemann sheet.

Al being the contour of integration shown in Figs. 5.1b,c when m / 0 and in Figs. 5.2b,c when m = 0. According to the connection formula (4.38a,b) for a real potential barrier the particular solution of the differential equation (2.32a,b), which to the left of the barrier is given... 

When x < t, analogous reasoning shows that the contour of integration must be closed to the left, and the simple pole at s = 0 and branch points at s = — 1 and at s = — deolf foY will provide a nonzero contribution to V. Thus some disturbance always propagates with the velocity... 

We will make use of Cauchy s theorem, according to which the integral value of an analytical function does not change under deformation of an integration contour if it does not intersect singularities on the complex plane of variable to. It is clear that deforming the contour of integration in the upper half-plane (Im m > 0) exponent e with an increase of Im TO tends to zero. 

Thus the contour of integration along the real axis of m is replaced by that along both sides of two cross-sections, where Re mi =0 and Re m2 = 0, respectively, and within an area surrounding the real axis of m and these contours singularities are absent (Fig. 4.14). 

Now we will derive expressions for the vertical component of the field on the borehole axis when there is an invasion zone and measurements are performed at the far zone. Taking into account that the integrand in eq. 4.136 is an even function we will consider integration along whole axis m and, applying Cauchy s theorem, the contour of integration then will be deformed in the upper part of the complex plane of m without intersection of singularities on this plane. 

Now we will obtain asymptotic formulae for the field in the far zone (a 1). In deriving a formula we will deform the contour of integration in eq. 10.33 on the complex plane of variable m. However, such a procedure requires either the proof of absence of poles of the integrand or evaluation of their contribution to the integral value. The problem of determination of poles is extremely difficult because of the complexity of the integrand. At the same time sufficient agreement of results of calculations by asymptotic and exact formulae allows us to think that if there are poles in the upper half-plane of m, their contribution in a considered part of the spectrum is sufficiently small. Let us present integral in eq. 10.33 in the form ... 

To find a suitable contour of integration, we first notice that a and c are both positive we have also from (6), that 6 > 0. This corresponds,... 

The second term in the braces vanishes if the poles of X (co), X/oo), Xt(a)) lie infinitesimally close to the imaginary frequency axis. The first term remains and, by shifting the contour of integration to the imaginary axis analogous to our procedure in the preceding section, becomes... 

---