I can write, $\int\limits_0^1 t^a(1/2 - t)^b\ dt = 2^{-(a+b)}\int_0^1 (2t)^a(1 - 2t)^b\ dt$.
Now, to compute $\int\limits_0^1 (2t)^a(1 - 2t)^b\ dt$, I can perform a substitution: compose the integrand with the function $u \mapsto u/2$ to get $\int_0^1 (2t)^a(1 - 2t)^b\ dt = \frac{1}{2}\int_0^2 u^a(1 - u)^b\ du$.
But what is $\int\limits_0^2 u^a(1 - u)^b\ du$? Is this some kind of Beta integral?
Making the problem more general $$\int\limits_0^1 t^a(c - t)^b\, dt \qquad \text{with} \qquad 0< c \leq 1$$ no problem with the antiderivative $$\int t^a(c - t)^b\, dt =c^{a+b+1} B_{\frac{t}{c}}(a+1,b+1)$$ No problem with $$I=\int\limits_0^c t^a(c - t)^b\, dt=c^{a+b+1}\frac{\Gamma (a+1)\, \Gamma (b+1) }{\Gamma (a+b+2)}\quad \text{if}\quad \Re(b)>-1\land \Re(a)>-1$$
If $ \Re(b)>-1\land \Re(a)>-1$ $$\int\limits_c^1 t^a(c - t)^b\, dt=(-1)^b c^{a+b+1}\frac{\Gamma (b+1) \Gamma (-a-b-1)-\Gamma (-a) B_c(-a-b-1,b+1)}{\Gamma (-a)}$$ and the problem comes from the $\color{red}{(-1)^b}$ term.
For illustartion, using $c=\frac 1 2$, $a=2$ and $b=3+\frac k{10}$, the result is $$I(k)=\frac{250- (-1)^{k/10}\, 5(k^2+100k+2450) }{2^{3+\frac{k}{10}}\, (k+40) (k+50) (k+60) }$$ and the values are $$\left( \begin{array}{cc} k & \text{numerical result} \\ 0 & -0.012500\, \\ 1 & -0.010863\,-0.003604 \, i \\ 2 & -0.008427\,-0.006268 \, i \\ 3 & -0.005557\,-0.007892 \, i \\ 4 & -0.002603\,-0.008490 \, i \\ 5 & +0.000137\,-0.008173 \, i \\ 6 & +0.002434\,-0.007118 \, i \\ 7 & +0.004138\,-0.005547 \, i \\ 8 & +0.005179\,-0.003694 \, i \\ 9 & +0.005563\,-0.001780 \, i \\ 10 & +0.005357\, \end{array} \right)$$
the numerical values are