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We have integral: $$\int_0^1\exp\left(n\left(\frac{itz}{\sqrt{a(1-a)n}}+a\ln(z)+(1-a)\ln(1-z)\right)\right)dz=\int_0^1\exp(nf(z))dz,$$ where $0<a<1$.

We want to approximate this integral when $n\rightarrow\infty$ by method of steepest descent. Why deforming $[0,1]$ to the contour through saddle point has the following expansion: $$z=a+\frac{it\sqrt{a(1-a)}}{\sqrt{n}}+O\left(\frac{1}{n}\right)?$$ I found saddle point $z_0=a$ from $\text{Re}'(f(z))=0$. I don't know what to do next? I can’t understand how they got it? I hope for your help! Thank you!

CROCO
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  • So $f(z)$ depends on $n$? – Qmechanic Jun 08 '20 at 08:30
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    This is a case which in general gives Faxen's integral (see Olver, *Asymptotics and Special Functions*, Ch. 9, Sec. 4), because $\psi(z)$ in $\exp(n \phi(z) + \psi(z) \sqrt n)$ cannot be replaced with $\psi(a)$ even though the integral $I$ is asymptotically equivalent to $\int_{a - \epsilon}^{a + \epsilon}$. We have $$I = (1 + o(1)) \int_{\mathbb R} \exp(n (\phi(a) + \phi''(a) (z - a)^2/2) + \psi(z) \sqrt n) \, dz.$$ Since $\psi$ is linear, the integral on the rhs is elementary. – Maxim Apr 03 '21 at 04:08

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my advice would be to try rewriting your integral as $$ I = \int_0^1 z^{an}(1-z)^{n(1-a)} \exp\left({\sqrt{n}xiz}\right) dz, $$ where $x(t,a):=\frac{t}{\sqrt{a(1-a)}}.$

Then there are no stationary points, but you have a simple linear phase, so steepest descent can be applied easily by deforming onto the vertical contours Re(z)=0 and Re(z)=1, like so $$ I = i\int_0^\infty (ip/x)^{an}(1-p/x)^{n(1-a)} \exp\left(-np\right) dp -ie^{\sqrt{n}xi}\int_0^\infty (1+ip/x)^{an}(p/x)^{n(1-a)} \exp\left(-np\right) dp $$ where the first and second integral correspond to the integral up Re(z)=0 and down Re(z)=1 respectively.

At this point, you could binomially expand the (1-p/x)^{n(1-a)} and (1+ip/x)^{an} terms, aiming for an asymptotic expansion of the integrand which holds for $p\to0$. Once you've done this, apply Watson's Lemma to both integrals.

I hope this helps :-)

AGibbs
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  • $I$ is asymptotically equivalent to $\int_{a - \epsilon}^{a + \epsilon}$, not $\int_0^{i \epsilon} + \int_{1 + i \epsilon}^1$. – Maxim Apr 03 '21 at 04:00