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Is there a closed form expression( special functions are allowed) for the following integral:

$$J(a,b)=\int_{0}^{0.5} e^{-0.5\frac{a}{2u-1}} \left(\frac{u}{1-u} \right)^{2b-1} \,du$$ where $a<0$ and $b\in (0,1)$.

Simulation studies showed that $J(a,b)$ exists.

#R code
a<<--1
b<<-0.5
fn<-function(u){
  exp(-0.5*a/(2*u-1))*(u/(1-u))^(2*b-1)
}
integrate(fn,lower=.01,upper=0.49)
#0.1572873 with absolute error < 1.2e-05
Bumblebee
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Masoud
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  • What's the question and may you provide an attempt? – Accelerator Apr 18 '23 at 22:32
  • 2
    For $b=\frac{1}{2}$ there exists a closed form $$J\left(a,\frac{1}{2}\right) \mapsto \frac{1}{4} \left(2 e^{a/2}-a\ Ei\left(\frac{a}{2}\right)\right)$$ with $Ei(z)$ the exponential integral function. – gpmath Apr 19 '23 at 06:28

1 Answers1

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For various ranges of $a$ and $b$ the integrand looks like a bell curve with a steeper slope around small values of $u$. The integrand rises as $u$ approaches $\tfrac{1}{4}$ and reaches very small values around $u=\tfrac{1}{2}$.

On approximating $\left(\dfrac{u}{1-u}\right)^{2b-1}$ by its expansion around the point $u=\tfrac{1}{4}$, a rough approximation to the maximum, I get, to the first order in $u-\tfrac{1}{4}$,

$$\left(\dfrac{u}{1-u}\right)^{2b-1}\approx 3/(3^b)^2 + ((32 b - 16) (u - 1/4))/(3^b)^2.$$ When $b=\tfrac{1}{2}$, this term reduces to unity, and the integral reduces to the form given by gpmath.

Inserting this expression into the integrand and using Wolfram Alpha produces the result $$\int \exp\left\{(-a/(2 (-1 + 2 u))\right\} \left\{3/(3^b)^2 + ((32 b - 16) (u - 1/4))/(3^b)^2\right\}\,{\rm d}u = (\tfrac{1}{4} 9^{-b} (2 (2 u - 1) e^{a/(2 - 4 u)} (a (2 - 4 b) + 8 (2 b - 1) u + 3) - a (a (4 b - 2) - 8 b + 1) \text{Ei}(a/(2 - 4 u))) + \text{constant}.$$

Differencing this relation for $u=\tfrac{1}{2}$ and $u=0$ would give a rough approximation to the integral. More terms in the Taylor expansion about the point $u=\tfrac{1}{4}$ could be taken, but the algebra increases exponentially, although I tried terms up to $\left(u-\tfrac{1}{4}\right)^3$ with no problems.