This page presents typical integrals appearing in dispersion relations as well as visualizations of their behavior on the complex plane. Taking the threshold, the lower limit of integration, to be unity for convenience, single dispersion relation integrals are of the general form

$I(s) =∫1∞ ds′ f(s′) s′-s$

where the argument s is energy squared and the function f is the discontinuity across the branch cut of the physical sheet.

The simplest such integral takes the discontinuity to be a constant. Since the resulting integral is divergent, a cutoff is needed. The integral is then

$I0(s) =∫1 sc ds′ s′-s =ln(s′ -s) | 1sc =ln( sc-s 1-s)$

This integral has two branch points, one at unity and the other at the cutoff. The first is visible by default in the following interactive graphic, and the other can be brought into view by lowering the cutoff value:

Changing the value of the cutoff does not appreciably alter the overall appearance of the surface, apart from its scale and the location of the second branch point as noted. The change in phase coloration as the cutoff is increased corresponds to the increased distance to the second branch point.

Another way to keep a dispersion integral convergent is by subtraction of the value of the integral at some point on the complex plane. This first subtraction is

$I0(1) (s) =I0(s) -I0 (s0) =∫1∞ ds′ [1 s′-s -1s′ -s0] I0(1) (s) =(s -s0) ∫1∞ ds′ (s′-s) (s′ -s0) =ln( 1-s0 1-s)$

This result now has only a single finite branch point. The visualization of this integral allows input of the evaluation point as a string for convenience:

The choice of evaluation point clearly allows rotation of the branch cut into any direction, as well as setting the overall scale of the phase structure. The evaluation point should be pure real to maintain the meaning of the branch cut of the physical sheet.

If the discontinuity function is linear the integral is again divergent. With an explicit cutoff the integral is

$I1(s) =∫1 sc ds′ s′ s′-s =s′ +sln (s′-s) |1 sc I1 (s) =sc-1 +sln( sc -s1-s )$

The factor on the logarithm needs to stay outside to avoid numerical clipping by the argument function. The integral can be visualized as before:

The second branch point can again be brought into view by lowering the cutoff value.

A single subtraction

$I1(1) (s) =I1(s) -I1 (s0) =(s-s0) ∫1∞ ds′ s′ (s′-s) (s′ -s0)$

is no longer enough to make this integral convergent, so define a higher subtraction:

$I1(2) (s) =I1(1) (s) -(s -s0) ∂∂s0 I1(s0) I1(2) (s) =(s-s0) ∫1∞ ds′ [s′ (s′-s) (s′ -s0) -s′ (s′ -s0 )2] I1(2) (s) =(s-s0 )2 ∫1∞ ds′ s′ (s′-s) (s′-s0 )2$

The integration is performed using partial fraction decomposition of the second step:

$I1(2) (s) =∫1∞ ds′ [s s′-s -s0 s′ -s0 -(s-s0) [1s′ -s0 +s0 (s′ -s0 )2]] I1(2) (s) =∫1∞ ds′ [s s′-s -s s′ -s0 -s0 s-s0 (s′ -s0 )2] I1(2) (s) =sln( 1-s0 1-s) -s0 s-s0 1-s0$

The visualization of this integral is

which differs principally from the visualization of the first subtraction for a constant discontinuity in its linear dependence on energy and doubled phase structure.

To account for higher powers in the numerator, one can define additional subtractions in the same way,

$Ik (n+1) (s) =Ik(n) (s) -(s-s0 )n n! ∂n ∂s0n Ik(s0) Ik (n+1) (s) =(s-s0 )n ∫1∞ ds′ [s′k (s′-s) (s′ -s0 )n -s′k (s′ -s0 )n+1] Ik (n+1) (s) =(s-s0 )n+1 ∫1∞ ds′ s′k (s′-s) (s′-s0 )n+1$

although such integrals will of course not converge until $n+1>k$ . Applying this term by term to the expansion of a general discontinuity function, one can immediately write an n-subtracted integral as

$If(n) (s) =(s-s0 )n ∫1∞ ds′ f(s′) (s′-s) (s′-s0 )n$

with the same caveat about convergence. With respect to the original integral, dispersion relations used phenomenologically are often written in the form

$I(s) =∑k=0 n-1 cksk +(s-s0 )n ∫1∞ ds′ fR (s′) (s′-s) (s′-s0 )n +(s-s0 )n ∫−1 −∞ ds′ fL (s′) (s′-s) (s′-s0 )n$

where a left-hand branch cut has been included for generality. This form can be derived by applying the Cauchy integral formula to the combination $I\left(s\right)/\left(s-{s}_{0}{\right)}^{n}$ .

In the interest of analytic behavior, consider an integral with an arbitrary power, which can be evaluated using a binomial integral. The value of the arbitrary power will be taken to be less than or equal to unity. First with an explicit cutoff one has

$Ia(s) =∫1 sc ds′ s′a s′-s =−s′ (a+1) s(a+1) F12 (1,a+1; a+2; s′s) |1 sc Ia (s) =1 s(a+1) [ F12 (1,a+1; a+2; 1s) -sc a+1 F12 (1,a+1; a+2; scs) ]$

which looks complicated, but can be easily visualized with Math:

This integral naturally agrees with previous results when $a\to 0$ or $a\to 1$ , since those previous integrals could have been evaluated as well using the binomial integral.

Now consider a single subtraction. Evaluations of the component integrals can proceed as for those with the cutoff, but it is not at all clear how they behave with respect to one another at the upper limit. The situation becomes clearer if that point is brought to the origin with the change of variable $w′ =1s′$ . The integral with a single subtraction is then

$Ia(1) (s) =Ia(s) -Ia (s0) =∫1∞ ds′ [s′a s′-s -s′a s′ -s0] Ia(1) (s) =∫01 dw′ [w′ −a-1 1-sw′ -w′ −a-1 1-s0 w′] Ia(1) (s) =−w′ −aa F12 (1,−a; 1-a; sw′) +w′ −aa F12 (1,−a; 1-a; s0w′) |01 Ia(1) (s) =1a [ F12 (1,−a; 1-a; s0) -F12 (1,−a; 1-a; s)]$

At the lower limit the leading terms of the hypergeometric functions cancel, and their second terms remove the singularity there as long as $a<1$ . The visualization of this integral is

Since the leading terms of the hypergeometric functions cancel, the integral with a single subtraction remains finite as $a\to 0$ . This can be understood by ignoring this parameter with respect to unity and evaluating

$lima→0 F12 (1,−a; 1;s) -1a =lima→0 ∑k=1 ∞ Γ(−a+k) aΓ(−a) skk! =−lima→0 ∑k=1 ∞ Γ(−a+k) Γ(1-a) skk! =−∑k=1 ∞ Γ(k) skk! =−∑k=1 ∞ skk =ln(1-s)$

Applying this result to the last expression gives the same result as for a constant discontinuity function with a single subtraction, so this is the correct way to evaluate the limit.

Unfortunately as $a\to 1$ the expression above does not stay finite. Here is the behavior of the integral as a function of a, with the real part in blue and imaginary in red:

Presumably this is because a single subtraction is not enough to ensure convergence, so consider a second subtraction:

$Ia(2) (s) =Ia(1) (s) -(s -s0) ∂∂s0 Ia(s0) Ia(2) (s) =Ia(1) (s) -(s -s0) ∫1∞ ds′ s′a (s′ -s0 )2 Ia(2) (s) =Ia(1) (s) -(s -s0) ∫01 dw′ w′−a (1-s0 w′ )2 Ia(2) (s) =Ia(1) (s) -(s -s0) w′ 1-a 1-a F12 (2,1-a; 2-a; s0w′) |01 Ia(2) (s) =Ia(1) (s) -s-s0 1-a F12 (2,1-a; 2-a; s0)$

The visualization of this doubly subtracted integral is

This integral has the same doubled phase structure as the integral with a linear discontinuity function. Presumably this is because both are double subtracted integrals. This integral remains finite as a function of a, again with the real part in blue and imaginary in red:

As $a\to 0$ the additional term in the doubly subtracted integral remains finite, providing an offset to the singly subtracted integral as a function of energy variables. For the limit $a\to 0$ write out the singular terms explicitly:

$1a[ F12 (1,−a; 1-a; s0) -F12 (1,−a; 1-a; s)] -s-s0 1-a F12 (2,1-a; 2-a; s0) =1a -s0 1-a +∑k=2 ∞ Γ(k+1) Γ(−a +k) aΓ(−a) Γ(1-a) Γ(1-a +k) s0k k! -1a +s1-a -∑k=2 ∞ Γ(k+1) Γ(−a +k) aΓ(−a) Γ(1-a) Γ(1-a +k) skk! -s-s0 1-a -(s-s0) ∑k=1 ∞ Γ(k+2) Γ(1-a +k) (1-a) Γ(1-a) Γ(2-a) Γ(2-a +k) s0k k! =−∑ k=2∞ s0k k-a +∑ k=2∞ skk-a -(s-s0) ∑k=1 ∞ k+1 k+1-a s0k$

One can now safely set $a=1$ and shift summation indices to give

$−s0 ∑k=1 ∞ s0kk +s∑ k=1∞ skk -(s-s0) ∑k=1 ∞ (1+1k) s0k =sln( 1-s0) -sln( 1-s) -(s-s0) s0 1-s0$

which as expected is the same result as for the integral with a linear discontinuity function.

It is perhaps interesting to compare the behavior of the singly subtracted and doubly subtracted integrals. The former has the real part in blue and and imaginary in red as before, while the latter has the real part in purple and the imaginary in magenta:

For some input parameters their relative behavior is similar, but this can change radically with other parameters. As pointed out earlier the doubly subtracted integral has a finite offset that depends on energy variables, and this is clear in the graphic. It remains a question as to the relative applicability of the two integrals.

Since the Foissart bound includes the square of a logarithm, one could consider dispersion relation integrals with logarithmic discontinuity functions. The results include polylogarithms, which are available in Math. This will be left as a future possible addition.