I am currently working on a problem and there is a specific type of integral appearing which looks somewhat close to the Beta function
$$ \int_0^1 t^{-\eta t} (1-t)^{-\eta (1-t)}dt,\quad \eta \geq 0 $$
As you see it can be thought of as a Beta Function $$ \int_0^1 t^{a} (1-t)^{b}dt, $$ where however the exponents $a$ and $b$ are now functions of $t$ itself
\begin{align*} a(t) &= -\eta t \\ b(t) &= -\eta (1-t). \end{align*}
Does anyone know whether this function has already been investigated under some name?
Thanks in advance!
I faced a very similar problem during my thesis work (that is to say $55$ years ago). There is no name for this integral.
Adapting to your problem, repeating what I did by hand, what is did was an expansion of the integrand around $t=\frac 12$ to have $$ t^{-\eta t}\,(1-t)^{-\eta (1-t)}=2^\eta\left(1+\eta \,\sum_{n=1}^\infty a_n \left(t-\frac{1}{2}\right)^{2n} \right)$$ where the $a_n$ are polynomials of degree $(n-1)$ in $\eta$. The first terms are given below $$\left( \begin{array}{cc} n & a_n \\ 1 & -2 \\ 2 & 2 \eta -\frac{4}{3} \\ 3 & -\frac{4 \eta ^2}{3}+\frac{8 \eta }{3}-\frac{32}{15} \\ 4 & \frac{2 \eta ^3}{3}-\frac{8 \eta ^2}{3}+\frac{232 \eta }{45}-\frac{32}{7} \\ 5 & -\frac{4 \eta ^4}{15}+\frac{16 \eta ^3}{9}-\frac{272 \eta ^2}{45}+\frac{3776 \eta }{315}-\frac{512}{45} \\ 6 & \frac{4 \eta ^5}{45}-\frac{8 \eta ^4}{9}+\frac{208 \eta ^3}{45}-\frac{43168 \eta ^2}{2835}+\frac{49024 \eta }{1575}-\frac{1024}{33} \\ 7 & -\frac{8 \eta ^6}{315}+\frac{16 \eta ^5}{45}-\frac{352 \eta ^4}{135}+\frac{35648 \eta ^3}{2835}-\frac{195584 \eta ^2}{4725}+\frac{904192 \eta }{10395}-\frac{8192}{91} \\ 8 & \frac{2 \eta ^7}{315}-\frac{16 \eta ^6}{135}+\frac{784 \eta ^5}{675}-\frac{21632 \eta ^4}{2835}+\frac{1523936 \eta ^3}{42525}-\frac{6185216 \eta ^2}{51975}+\frac{1211468288 \eta }{4729725}-\frac{4096}{15}\\ 9 & -\frac{4 \eta ^8}{2835}+\frac{32 \eta ^7}{945}-\frac{32 \eta ^6}{75}+\frac{30976 \eta ^5}{8505}-\frac{969664 \eta ^4}{42525}+\frac{49704448 \eta ^3}{467775}-\frac{25325513728 \eta ^2}{70945875}+\frac{35332096 \eta }{45045}-\frac{131072}{153} \end{array} \right)$$
Which gives, as an approximation, $$ \int_0^1t^{-\eta t}\,(1-t)^{-\eta (1-t)}\,dt=2^\eta\left(1+\eta \,\sum_{n=1}^\infty \frac{2^{-2 n} }{2 n+1}a_n \right)$$
Using these terms, the definite integral write $$ \int_0^1t^{-\eta t}\,(1-t)^{-\eta (1-t)}\,dt=2^\eta\,\left(1+\sum_{n=1}^9 b_n \eta^n\right)$$ where the $b_n$' are $$\left\{-\frac{8960447}{46558512},\frac{1098559307}{30875644800},-\frac{7777474493}{ 1466593128000},\frac{197620627}{309435033600},-\frac{71722879}{1149330124800}, \frac{3576397}{731391897600},-\frac{1231}{4167475200},\frac{25}{2000388096},-\frac {1}{3530096640}\right\} $$ Some results $$\left( \begin{array}{ccc} \eta & \text{approximation} & \text{exact} \\ 0.0 & 1.00000 & 1.00000 \\ 0.5 & 1.28982 & 1.28950 \\ 1.0 & 1.67680 & 1.67621 \\ 1.5 & 2.19566 & 2.19483 \\ 2.0 & 2.89394 & 2.89291 \\ 2.5 & 3.83682 & 3.83562 \\ 3.0 & 5.11386 & 5.11252 \\ 3.5 & 6.84820 & 6.84674 \\ 4.0 & 9.20942 & 9.20785 \\ 4.5 & 12.4313 & 12.4296 \\ 5.0 & 16.8364 & 16.8347 \end{array} \right)$$