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Let $0< b\leqslant 1$. I am interested in using the steepest descent method to calculate the asymptotic approximation (as $x\to+\infty$) to the following integral that is related to function ${_1}F_1(1,9/4+ibx,-x)$:

$$J(b,x)=\int_0^{1}s^{1/4+i b x}e^{sx}ds=\int_0^{1}s^{1/4}\exp(x(s+ib\log s))ds\tag{1}$$

Let $\rho(z)=z+ib\log z$. From several excellent solutions in the posts here and here, I know that we suppose to find contour segments such that $\Im(\rho(z))$=const. I was not able to find explicit solutions when I set $z=u+iv$ or $z=re^{i\phi}$.

Also $\rho'(z)=1+ib/z$ so the saddle point is at $z_0=-ib$. And $\rho^{"}(z_0)=i/b$.

Thanks for any suggestions.

Update

Set $z=re^{i\phi}$ in $$\rho(z)=z+ib\log z,$$ we obtain $$\Re\rho(z)=r\cos\phi-b\phi,\quad\Im\rho(z)=r\sin\phi+b\log r$$ To make $\Im\rho(z)=-c$ we can set $$\sin\phi=-\frac{b\log r}{r}-\frac{c}{r},\qquad\phi=-\arcsin\left(\frac{b\log r}{r}+\frac{c}{r}\right)$$

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    you can expand in a $\delta$-neighborhood of the critical points which are the endpoints $z=0,1$ and the saddlepoint at $z=-ib$ so you don't need explicit formulas for the paths of steepest descent – 2017-02-15
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    furthermore, ask yourself the question in which sector of the complex plane $f(z,x)= z^{1/4}e^{x\rho(z)}$ converges – 2017-02-15
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    to 0 for large z – 2017-02-15
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    @tired Thanks for the comments. I will try what you suggested. – 2017-02-16

1 Answers 1

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If you start with the ${_1\hspace{-2px}F_1}$ representation $$J(b, x) = \frac {e^x} {\frac 5 4 + i b x} \,{_1\hspace{-2px}F_1} {\left( 1, \frac 9 4 + i b x, -x \right)},$$ then $$ \lim_{x \to \infty} {_1\hspace{-2px}F_1} {\left( 1, \frac 9 4 + i b x, -x \right)} = \sum_{k = 0}^\infty \lim_{x \to \infty} \frac {(-x)^k} {\left( \frac 9 4 + i b x \right)_k} = \sum_{k = 0}^\infty \left( \frac i b \right)^k = \frac b {b - i}$$ and $$J(b, x) \sim \frac {e^x} {(i b + 1) x}.$$ This shows that the asymptotic is not determined by the saddle point at $-i b$.

The application of the steepest descent method would go as follows. The curve $\operatorname{Im} \rho(z) = \text{const}$ passing through $z = 1$ is $$\operatorname{Im} (\rho(u + i v) - \rho(1)) = v + \frac {b \ln(u^2 + v^2)} 2 = 0.$$ The slope at $z = 1$ is $-b$. Taking $z = 1 + (1 - i b)\xi$, we obtain $$\rho(1 + (1 - i b)\xi) = 1 + (1 + b^2) \xi + O(\xi^2), \\ (1 - i b) z^{1/4} e^{x \rho(z)} \Big\rvert_{z = 1} \int_{-\infty}^0 e^{(1 + b^2) x \xi} d\xi = \frac {e^x} {(i b + 1) x}.$$ It remains to show that other contributions are negligible, using the fact that in the region $\operatorname{Re} z < 1 \land \operatorname{Im} z > 0$, we have $$\operatorname{Re} \rho(z) = \operatorname{Re} z - b \arg z < \operatorname{Re} \rho(1).$$