Integrate solution doesn't work when values are substituted

Question

I would like a closed form for $$\text{Integrate}\left[\frac{(1-y)^{\frac{k}{2}+\frac{t}{2}-\frac{1}{2}} (y+1)^{\frac{t}{2}-\frac{k}{2}}}{\sqrt{1-y}-\sqrt{y+1}}-\frac{(1-y)^{\frac{k}{2}+\frac{t}{2}-1} (y+1)^{-\frac{k}{2}+\frac{t}{2}-\frac{1}{2}}}{\sqrt{1-y}-\sqrt{y+1}},\{y,0,1\},\text{Assumptions}\to \{k\in \mathbb{Z},t\in \mathbb{Z}\}\right]$$

Plotting the expression for k = 5, t = 10 say, and we see that this should be well defined. Mathematica gives us the solution $$-\frac{-2^{\frac{t}{2}+1} \Gamma \left(\frac{1}{2} (-k-t+3)\right) \left(\Gamma (t) \left(\pi \left(k^2-4 k+3\right) 2^{\frac{t+1}{2}} (-k+t+2) \csc \left(\frac{1}{2} \pi (k+t)\right) \Gamma \left(\frac{1}{2} (-k+t+2)\right)-2^{k/2} \Gamma (t+1) \left(\sqrt{2} \left(k^2-2 k (t+2)+t^2+4 t+3\right) \, _2F_1\left(\frac{1}{2} (-k-t+2),\frac{1}{2} (-k+t+2);\frac{1}{2} (-k+t+4);\frac{1}{2}\right)-(k-t-2) \left(2 (k-t-3) \, _2F_1\left(\frac{1}{2} (-k-t+3),\frac{1}{2} (-k+t+1);\frac{1}{2} (-k+t+3);\frac{1}{2}\right)+(-k+t+1) \, _2F_1\left(\frac{1}{2} (-k-t+3),\frac{1}{2} (-k+t+3);\frac{1}{2} (-k+t+5);\frac{1}{2}\right)\right)\right) \Gamma \left(-\frac{k}{2}-\frac{t}{2}+1\right)\right)+\pi 2^{\frac{t+1}{2}} \left(2 k^2-k (3 t+8)+t^2+6 t+8\right) \Gamma (t+1) \csc \left(\frac{1}{2} \pi (k+t)\right) \Gamma \left(\frac{1}{2} (-k+t+2)\right)\right)-\pi 2^{t+\frac{1}{2}} \left(k^2-2 k (t+2)+t^2+4 t+3\right) {\sec \left(\frac{1}{2} \pi (k+t)\right)} \Gamma \left(-\frac{k}{2}-\frac{t}{2}+1\right) \left(\left(-k^2+3 k-2\right) \Gamma (t)+(t+1) \Gamma (t+1)\right) \Gamma \left(\frac{1}{2} (-k+t+1)\right)}{4 \sqrt{2} (-k+t+1) (-k+t+2) (-k+t+3) \Gamma (t) \Gamma (t+1) \Gamma \left(\frac{1}{2} (-k-t+3)\right) \Gamma \left(-\frac{k}{2}-\frac{t}{2}+1\right)}$$ assuming $k + t > 0$. If we now try to substitute integer values of $k$ and $t$ into this, it gives an answer of Indeterminate.

How can I persuade Mathematica to give me an expression for the integral that works on integer values (I've tried to use Refine with integer constraints on the result, but that does not change the expression)

Answer 1

History of edits

3.7.18 Simplified derivation of closed form expression in terms of incomplete Beta functions

1.7.18 Complete solution provided

Summary

I have derived an "Integrate[] solution which works fine when values are substituted".

The expression h[k,t] for the integral obtained here is finite in the whole (k,t)-plane except for integer grid points (both k and t are integers) with k+t<=0 where singularities are encountered. These singularities are simple poles (rather than discontinuities as was stated by Bob Hanlon).

On the integers grid the singularities lie on the lines k+t=-n, n=0,1,2,...

Specifically, h[k,t] is finite for all integers k and t with k+t>0.

The behaviour of the integral, especially close to the singularities is illustrated graphically.

Simplified derivation

A compact derivation of a closed expression for the integral goes like this

(1) remove the denominator in the integrand

This is effectively multiplying denominator and numerator with (Sqrt[1 - y] + Sqrt[y + 1]).

f=((1 - y)^(k/2 + t/2 - 1/2) (y + 1)^(t/2 - k/2)/(Sqrt[1 - y] -
            Sqrt[y + 1]) - (1 - y)^(k/2 + t/2 - 
           1) (y + 1)^(-k/2 + t/2 - 1/2)/(Sqrt[1 - y] - Sqrt[y + 1]))

$$f = \frac{(1-y)^{\frac{1}{2} (k+t-1)} \left((y+1)^{\frac{t-k}{2}}-(y+1)^{\frac{1}{2} (-k+t-1)}\right)}{\sqrt{1-y}-\sqrt{y+1}}$$

f1 = FullSimplify[f]

1/2 (1 - y)^(1/2 (-3 + k + t)) (1 + y)^(
1/2 (-1 - k + t)) (-1 + y + Sqrt[1 - y^2]) 

$$f_{1}=\frac{1}{2} \left(\sqrt{1-y^2}+y-1\right) (1-y)^{\frac{1}{2} (k+t-3)} (y+1)^{\frac{1}{2} (-k+t-1)}$$

(2) Substitute y->1-2x (y=0..1 -> x=0..1/2)

f2 = FullSimplify[2 f1 /. y -> 1 - 2 x, 0 < x < 1/2] // FunctionExpand // 
  Expand

2^(-1 + t) (1 - x)^(-(k/2) + t/2) x^(-1 + k/2 + t/2) - 
 2^(-1 + t) (1 - x)^(-(1/2) - k/2 + t/2) x^(-(1/2) + k/2 + t/2)

$$f_{2}=2^{t-1} (1-x)^{\frac{t}{2}-\frac{k}{2}} x^{\frac{k}{2}+\frac{t}{2}-1}-2^{t-1} (1-x)^{-\frac{k}{2}+\frac{t}{2}-\frac{1}{2}} x^{\frac{k}{2}+\frac{t}{2}-\frac{1}{2}}$$

Apart from the factor $2^{t-1}$ each summand has the form $x^{a-1} (1-x)^{b-1}$ which when integrated between 0 and 1/2 gives the (incomplete) Euler Beta-function

Integrate[x^(a - 1) (1 - x)^(b - 1), {x, 0, 1/2}]

ConditionalExpression[Beta[1/2, a, b], Re[a] > 0]

Hence a closed form solution of the integral is

h0[k_, t_] = 
  2^(-1 + t) ((Beta[1/2, a, b] /. {a -> k/2 + t/2, 
        b -> 1 - k/2 + t/2}) - (Beta[1/2, a, b] /. {a -> 1 - 1/2 + k/2 + t/2, 
        b -> 1 - 1/2 - k/2 + t/2}));

$$h_{0}(k,t) = 2^{t-1} \left(B_{\frac{1}{2}}\left(\frac{k}{2}+\frac{t}{2},-\frac{k}{2}+\frac{t}{2}+1\right)-B_{\frac{1}{2}}\left(\frac{k}{2}+\frac{t}{2}+\frac{1}{2},-\frac{k}{2}+\frac{t}{2}+\frac{1}{2}\right)\right)$$

Inserting values gives

h0[5, 10]

512 (-(7/46080) + Beta[1/2, 15/2, 7/2])

numerically

N[%, 20]

0.011734378351166407572

The singularity structure is now easy to see from the fact that $B_{\frac{1}{2}}(a,b)$ has simple poles at negative integer values of the parameter "a" and is holomorph in "b".

Calculations

I do this "with a little help for my friend", Mathematica. The help consists in a simple substitution and a subsequent manual simplification.

$Version

"10.1.0  for Microsoft Windows (64-bit) (March 24, 2015)"

ClearAll[f, g, h, t, k, y, z]

The integrand in question is

f[k_, t_, y_] := ((1 - y)^(k/2 + t/2 - 1/2) (y + 1)^(t/2 - k/2)/(Sqrt[1 - y] -
        Sqrt[y + 1]) - (1 - y)^(k/2 + t/2 - 
       1) (y + 1)^(-k/2 + t/2 - 1/2)/(Sqrt[1 - y] - Sqrt[y + 1]))

The numerical value of the integral for parameters k=5, t=10 is

iN1 = NIntegrate[f[5, 10, y], {y, 0, 1}, WorkingPrecision -> 20]

0.011734378351166406412

Now our "first aid": the substitution 1-y -> z

g[k_, t_, z_] = 
 1/2 (1 - y)^(1/2 (-3 + k + t)) (1 + y)^(
    1/2 (-1 - k + t)) (-1 + y + Sqrt[1 - y^2]) /. y -> 1 - z // Simplify

1/2 (2 - z)^(1/2 (-1 - k + t)) z^(1/2 (-3 + k + t)) (-z + Sqrt[-(-2 + z) z])

Numerical check of the corresponding integral:

iN2 = NIntegrate[g[5, 10, z], {z, 0, 1}, WorkingPrecision -> 20]

0.011734378351166406412

iN1 == iN2

True

Now we do the integration without Assumptions[] on k and t which is performed by Mathematica in seconds:

i = Integrate[g[k, t, z], {z, 0, 1}]

ConditionalExpression[(1/((k + t) (1 + k + t)))
 2^(-1 - k) (2 E^(
     1/2 (k + t) (-I \[Pi] + Log[2])) (1 + k + t) Hypergeometric2F1[(k - t)/
      2, (k + t)/2, 1/2 (2 + k + t), 1/2] - 
    I E^(1/2 (1 + k + t) (-I \[Pi] + Log[2])) (k + t) Hypergeometric2F1[
      1/2 (1 + k - t), 1/2 (1 + k + t), 1/2 (3 + k + t), 1/2]) (Cos[
     1/2 \[Pi] (k + t)] + I Sin[1/2 \[Pi] (k + t)]), Re[k + t] > 0]

Keeping in mind the condition

Re[k+t]>0 (1)

we extract the expression

h[k_, t_] = i[[1]];

Remark: the indefinte integral is calculated as well (we skip the expression here for brevity)

Evaluation

Now we can substitute integer values for k and t respecting condition (1)

For the "standard" values of k and t we find

h[5, 10] // Simplify
% // N

1/90 (-7 + 6 Hypergeometric2F1[-(5/2), 1, 17/2, -1])

0.0117344

For the smallest integer parameters permitted by (1) we have

h[0, 1] // Simplify

\[Pi]/4

h[1, 0] // Simplify

1/4 (\[Pi] - Log[4])

Going further up on the (k,t)-integer grid:

First on the "borders"

Table[h[0, t], {t, 1, 5}] // Simplify

{\[Pi]/4, (6 - \[Pi])/8, -(1/6) + \[Pi]/8, 
 1/96 (44 - 9 \[Pi]), -(1/6) + (3 \[Pi])/32}

Table[h[k, 0], {k, 1, 5}] // Simplify

{1/4 (\[Pi] - Log[4]), 1/4 (-4 + \[Pi] + Log[4]), 1/4 (2 - \[Pi] + Log[4]), 
 1/12 (14 - 3 \[Pi] - 6 Log[2]), 1/12 (-5 + 3 \[Pi] - 3 Log[4])}

Then on the diagonal

Table[h[n, n], {n, 1, 5}] // Simplify

{1/6 (3 - 2 Hypergeometric2F1[1/2, 1, 5/2, -1]), 
 1/4 - 1/5 Hypergeometric2F1[1/2, 1, 7/2, -1], 
 1/6 - 1/7 Hypergeometric2F1[1/2, 1, 9/2, -1], 
 1/8 - 1/9 Hypergeometric2F1[1/2, 1, 11/2, -1], 
 1/10 - 1/11 Hypergeometric2F1[1/2, 1, 13/2, -1]}

Also there is no problem with real values for k and t

Numerical example

h[8.5, 3.5]

0.00661082 + 9.56017*10^-17 I

Symbolic example

h[\[Pi]/4, E] // Simplify

(1/((4 E + \[Pi]) (4 + 
   4 E + \[Pi])))4 (4 Hypergeometric2F1[1, 1/8 (-4 E + \[Pi]), 
     1/8 (8 + 4 E + \[Pi]), -1] + 
   4 E (Hypergeometric2F1[1, 1/8 (-4 E + \[Pi]), 1/8 (8 + 4 E + \[Pi]), -1] - 
      Hypergeometric2F1[1, 1/8 (4 - 4 E + \[Pi]), 
       1/8 (12 + 4 E + \[Pi]), -1]) + \[Pi] (Hypergeometric2F1[1, 
       1/8 (-4 E + \[Pi]), 1/8 (8 + 4 E + \[Pi]), -1] - 
      Hypergeometric2F1[1, 1/8 (4 - 4 E + \[Pi]), 
       1/8 (12 + 4 E + \[Pi]), -1]))

Here's an example with one negative integer k and one non integer t violating the condition k+t>0 which results in a finite value, as it should be:

With[{k = -4, t = \[Pi]}, Print[k + t // N]; h[\[Pi], -4] // Simplify]

 -0.858407

(54 - 36 \[Pi] + 25 \[Pi]^2 + 21 \[Pi]^3 - 8 \[Pi]^4 - 
 3 \[Pi]^5 + \[Pi]^6)/(2 (-4 + \[Pi]) (-3 + \[Pi]) \[Pi] (-4 + \[Pi]^2) (-3 - \
\[Pi] + 3 \[Pi]^2 + \[Pi]^3))

Singularity structure

The overall behaviour of h in the real (k,t)-plane is shown in this 3D graph

We identify lines of singularity given by

k+t=-n, n = 0,1,2,... 

The next graph shows h[k,t] along the line t=k, i.e. perpendicular to the lines of singularity

The cross section shows that h[k,k] has poles at k = -(2m+1), m=0, 1, 2, ...

The next two graphs follow the pole lines k+t==-n+[Epsilon] in an [Epsilon]-vicinity for n=0 and n=4 respectively:

We have plotted h*[Epsilon] which is almost independent of [Epsilon]. This shows that h diverges like 1/[Epsilon] for [Epsilon]->0.

Note added on 2.7.

In oder to study the singularities in more detail it is convenient to take k+t=m as the parameter besides k:

hm[k_,m_] = h[k, t] /. t -> m - k // Simplify

(1/(m (1 + m)))2^(-1 - k + m/
  2) (2 (1 + m) Hypergeometric2F1[k - m/2, m/2, 1 + m/2, 1/2] - 
   Sqrt[2] m Hypergeometric2F1[1/2 + k - m/2, (1 + m)/2, (3 + m)/2, 1/2])

This also simplifies the expression further.

For given k this function has simple poles in m for negative integers. This can be derived from the hypergeometric functions. But we prefer to illustrate it graphically (k can be any real number, even integers; we make two specific choices)

Positive k:

With[{k = 5}, Plot[hm[k,m], {m, -15, 2}, PlotRange -> {-5, 5}]]

Negative k:

With[{k = -5}, Plot[hm[k,m], {m, -15, 2}, PlotRange -> {-1, 1}]]

In the case of k<0 the |k| poles from -|k|-1 to -2|k| vanish.

Answer 2

$Version

(* "11.3.0 for Mac OS X x86 (64-bit) (March 7, 2018)" *)

Clear[f]

f[k_, t_] = Assuming[Element[{k, t}, Integers] && k + t > 0,
  Integrate[(1 - y)^(k/2 + t/2 - 1/2) (y + 1)^(t/2 - k/2)/(Sqrt[1 - y] - 
        Sqrt[y + 1]) - (1 - y)^(k/2 + t/2 - 
        1) (y + 1)^(-k/2 + t/2 - 1/2)/(Sqrt[1 - y] - Sqrt[y + 1]), {y, 0, 1}]]

(* -((2^(-1 + t) Gamma[1/2 (1 - k + t)] Gamma[1/2 (-1 + k + t)])/Gamma[t]) + (
 2^(-1 + t) Gamma[1/2 (3 - k + t)] Gamma[1/2 (-1 + k + t)])/Gamma[1 + t] + (
 2^(-1 + t) Gamma[1/2 (2 - k + t)] Gamma[(k + t)/2])/
 Gamma[1 + t] + (2^(
    1/2 (-4 + k + 
       t)) (2 (3 + k^2 + 4 t + t^2 - 2 k (2 + t)) Hypergeometric2F1[
        1/2 (2 - k - t), 1/2 (2 - k + t), 1/2 (4 - k + t), 1/2] - 
      Sqrt[2] (-2 + k - 
         t) (2 (-3 + k - t) Hypergeometric2F1[1/2 (3 - k - t), 
           1/2 (1 - k + t), 1/2 (3 - k + t), 1/
           2] + (1 - k + t) Hypergeometric2F1[1/2 (3 - k - t), 
           1/2 (3 - k + t), 1/2 (5 - k + t), 1/2])))/((-3 + k - t) (-2 + k - 
      t) (-1 + k - t)) *)

Plotting the function shows that it is discontinuous along stripes (t - k == n, Element[n, Integers] && n < 0

Plot3D[f[k, t], {k, 0, 10}, {t, 0, 10}, 
 AxesLabel -> (Style[#, 14, Bold] & /@ {"k", "t"}), MaxRecursion -> 3, 
 PlotPoints -> 50, ClippingStyle -> None, Mesh -> {9, 9}]

f[5, 10]

(* -(64/45) + (143 \[Pi])/512 - 
 2/21 Sqrt[2] (-(847/(60 Sqrt[2])) + (11 (2048 + 819 \[Pi]))/(6144 Sqrt[2])) *)

% // N

(* 0.0117344 *)

f[8.5, 3.5] // Quiet

(* Indeterminate *)

max = 8;
(Table[f[k, t], {k, 0, max}, {t, 0, max - 1}] // N) /. {ComplexInfinity -> 
     "*", Indeterminate -> "--"} // 
  TableForm[#, TableHeadings -> {Range[0, max], Range[0, max]}] & // Quiet

You may be able to evaluate at the discontinuities by taking the limit

Limit[f[5, t] // Simplify, t -> 3] // N

(* 0.0170566 *)