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.