In Appendix A of [this paper](https://arxiv.org/pdf/cond-mat/9904278.pdf), it is claimed that the asymptotic behaviour of


$$\phi\_1(y,\lambda)=\frac{1}{\Gamma(\frac{1-\lambda}{2})}\int\_0^\infty dt~e^{-t}\cos(2y\sqrt{t})t^{-\frac{1+\lambda}{2}},$$


for $\lambda\_2=\text{Im}(\lambda)\to+\infty$ (where $\lambda=\lambda\_1+i\lambda\_2$), is


$$\phi\_1(y,\lambda\_1+i\lambda\_2)\sim\cosh[y\sqrt{\lambda\_2-i\lambda\_1}(1+i)]\exp(-y^2/2) .$$


I have been trying various asymptotic methods to arrive at this result, such as the method of stationary phase, without success. How might one derive this result?



Asymptotic behaviour of function from integral representation
If we expand $\cos{(2y\sqrt{t})}$ into Taylor series and integrate term by term, we get $$\phi\_1(y,\lambda)=\sum\limits\_{n=0}^\infty\frac{(-1)^n}{(2n)!}\frac{\Gamma\left(n+\frac{1}{2}-\frac{\lambda}{2}\right)}{\Gamma{\left(\frac{1}{2}-\frac{\lambda}{2}\right)}}(2y)^{2n}.$$ Now let us use the following asymptotic expansion of Tricomi and Erdelyi (see, for example, <https://www.sciencedirect.com/science/article/pii/S0022247X11011280> ) $$\frac{\Gamma(z+\alpha)}{\Gamma(z+\beta)}=\sum\_{n=O}^\infty C\_n(\alpha-\beta,\beta)z^{\alpha-\beta-n},$$ where $|z|\to\infty$ and the coefficients are given by reccurence relations (with $C\_0=1$) $$C\_n(\alpha-\beta,\beta)=\frac{1}{n}\sum\_{i=0}^{n-1}\left[\binom{\alpha-\beta-i}{n-i+1}-(-1)^{n+i}(\alpha-\beta)\beta^{n-i}\right]C\_i(\alpha-\beta,\beta).$$ As a result we get (asymptotically) $$\phi\_1(y,\lambda)=\sum\_{n=0}^\infty\sum\_{m=0}^\infty\frac{(-1)^n}{(2n)!}C\_m\left(n,1/2\right)\left(-\frac{\lambda}{2}\right)^{n-m}(2y)^{2n}.$$ It follows from the equation 1.4 of the mentioned paper that $C\_m(n,\beta)$ is proportional to $1/(n-m)!$. Therefore $C\_m(0,\beta)=C\_m(1,\beta)=\cdots=C\_m(m-1,\beta)=0$. As a result $$\phi\_1(y,\lambda)=\sum\_{m=0}^\infty\sum\_{n=m}^\infty\frac{(-1)^n}{(2n)!}C\_m(n,1/2)\left(-\frac{\lambda}{2}\right)^{n-m}(2y)^{2n}=\\ \sum\_{m=0}^\infty\sum\_{n=0}^\infty\frac{(-1)^{n+m}}{[2(n+m)]!}C\_m(n+m,1/2)\left(-\frac{\lambda}{2}\right)^n(2y)^{2n+2m}.$$ We can rewrite this as follows $$\phi\_1(y,\lambda)=\sum\_{m=0}^\infty\sum\_{n=0}^\infty\frac{(-1)^n}{(2n)!}\left(\sqrt{-\frac{\lambda}{2}}\;2y\right)^{2n}\frac{(2n)!}{[2(n+m)]!}C\_m(n+m,1/2)(-4y^2)^m.$$ As $|\lambda|\to\infty$, we can assume that for any fixed $m$ only terms $n\gg m$ are relevant (this is somewhat not rigorous. Can one make it more precise?). But then $$\frac{(2n)!}{[2(n+m)]!}\to\frac{1}{(2n)^{2m}}.$$ On the other hand, in the $n\gg m$ limit we get from the reccurence relations $$\frac{C\_m(n,1/2)}{n^{2m}}=\frac{1}{m}\frac{1}{n^{2m}}\left [\binom{n}{2}+\frac{n}{2}\right]C\_{m-1}(n,1/2)=\frac{1}{2m}\frac{C\_{m-1}(n,1/2)}{n^{2(m-1)}}.$$ This implies that $$\frac{C\_m(n,1/2)}{n^{2m}}\to \frac{1}{2^m\, m!}.$$ Therefore asymptotically $$\phi\_1(y,\lambda)=\sum\_{m=0}^\infty\sum\_{n=0}^\infty\frac{(-1)^n}{(2n)!}\left(\sqrt{-\frac{\lambda}{2}}\,2y\right)^{2n}\frac{1}{m!}\left(-\frac{y^2}{2}\right)^m=\cos{\left(\sqrt{-\frac{\lambda}{2}}\,2y\right)}e^{-y^2/2}.$$ To finish the proof, let us note that $$\sqrt{-\frac{\lambda}{2}}\,2y=iy\sqrt{\lambda\_2-i\lambda\_1}\,(1+i)$$ because $\sqrt{i}=(1+i)/\sqrt{2}$.


