10. Basic Integral for the de Sitter-de Sitter case.

The calculation of the integral (91) is performed in the complex plane by considering the function

$$f ( z ) = \sqrt{z} \sqrt{z - 1} \left( z - y_0 \right) \left( z - y_1 \right) ^{-1} \left( z - y_2 \right) ^{-1} \quad$$ (127)

which has two branch points (z=0 and z=1), and two simple poles (z=y₁ and z=y₂). Accordingly, we integrate along the path $\Gamma _{\epsilon _1 \, , \: \epsilon _2}$:

$C_{\epsilon _1 }(0)$ : a clockwise circumference of radius $\epsilon _1$ and center z=0;

$0+ \epsilon _1 \rightarrow 1 - \epsilon _2$: an oriented line segment of the x-axis in the upper half of the complex plane;

$C_{\epsilon _2 } (1)$: a clockwise circumference of radius $\epsilon _2$ and center z=1;

$0+ \epsilon _1 \leftarrow 1 - \epsilon _2$: an oriented line segment of the x-axis in the lower half of the complex plane

with

$$0 \leq \arg \left( z \right) < 2 \pi$$ (128)

$$0 \leq \arg \left( z - 1 \right) < 2 \pi \quad .$$ (129)

  

Figure: Integration path $\Gamma _{\epsilon _1 \, , \: \epsilon _2}$.

Then, we have

$$\lim _{\epsilon _1 \, , \: \epsilon _2 \to 0} \int _{\Gamm... ... {\cal R}{\rm {}es} \left\{ f , y_2 \right\} \right] \quad .$$ (130)

It is also true that

$$\lim _{\epsilon _1 \, , \: \epsilon _2 \to 0} \int _{\Gamma ... ...n _2}} dz f(z) = \int _0^1 dz f_+(z) - \int _0^1 dz f_-(z)$$ (131)

with

$$\! = \! \left [ \vert z \vert \right ] ^{\frac{1}{2} } \left [ \vert z - ... ...^{i \pi } \right ] ^{-1} \left [ \vert z - y _2 \vert e^{i \pi } \right ] ^{-1}$$ f ₊ ( z )

$$\! = \!$$ i f(y) (132)

$$\! = \! \left [ \vert z \vert e^{2 i \pi } \right ] ^{\frac{1}{2} } \left... ...^{i \pi } \right ] ^{-1} \left [ \vert z - y _2 \vert e^{i \pi } \right ] ^{-1}$$ f _- ( z )

$$\! = \! - i f(y) \quad .$$ (133)

Therefore,

$$\lim _{\epsilon _1 \, , \: \epsilon _2 \to 0} \int _{\Gamma _{\epsilon _1 \, , \: \epsilon _2}} dz f(z) = 2 i I$$ (134)

with

$$I = \pi \left [ {\cal R}{\rm {}es} \left\{ f , +\infty \righ... ... {\cal R}{\rm {}es} \left\{ f , y_2 \right\} \right ] \quad .$$ (135)

The residue of the function at infinity is found from its asymptotic expansion in powers of z^-1:

$$z^{\frac{1}{2} } \frac{1}{\left( z - 1 \right) ^{\frac{1}{2}}} \left( z - y_0 \right) \frac{1}{z - y_1} \frac{1}{z - y_2}$$ f(z) =

$$z ^{\frac{1}{2} } \frac{1} {\displaystyle z ^{\frac{1}{2}} \left(... ...{y_1}{z} \right) } \frac{1} {\displaystyle z \left( 1 - \frac{y_1}{z} \right) }$$ =

$$\sim \frac{1}{z} \left( 1 + \frac{1}{2z} + \frac{3}{8z ^{2}} + {\cal O} \left( z^{-3} \right) \right) \cdot$$

$$\qquad \qquad \qquad \cdot \left( 1 + \frac{y_1}{z} + \frac{y_1^{... ... \frac{y_2}{z} + \frac{y_2^{2}}{z^{2}} + {\cal O} \left( z^{-3} \right) \right)$$

$$\sim \frac{1}{z} + {\cal O} \left( \frac{1}{z^{2}} \right)$$ (136)

The opposite of the coefficient of z ^-1 yields the result:

$${\cal R}{\rm {}es} \left\{ f , +\infty \right\} = -1 \quad .$$ (137)

Next, the residues at the poles z=y₁ , y₂ are given by

$${\cal R}{\rm {}es} \left\{ f , y_1 \right\} = y_1^{\frac{1}... ...{2}} \left( y_1 - y_0 \right) \left( y_1 - y_2 \right) ^{-1}$$ (138)

$${\cal R}{\rm {}es} \left\{ f , y _2 \right\} = y_2^{\frac{1... ...eft( y_2 - y_0 \right) \left( y_2 - y_1 \right) ^{-1} \quad .$$ (139)

Then, summing up our results, we finally obtain the expression of the integral

$$\pi \left\{ \left( \frac{y_1 }{y_1 - 1} \right) ^{\frac{1}{2} } \left( y_1 - y_0 \right) \left( y_1 - y_2 \right) ^{-1} + \right.$$ I =

$$\qquad \qquad \qquad + \left. \left( \frac{y_2 }{y_2 - 1} \right)... ... } \left( y_2 - y_0 \right) \left( y_2 - y_1 \right) ^{-1} - 1 \right\} \quad .$$ (140)