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edited Mar 11, 2018 at 6:05

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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 ) $$\frac{\Gamma(z+\alpha)}{\Gamma(z+\beta)}=\sum_{n=O}^\infty C_n(\alpha-\beta,\beta)x^{\alpha-\beta-n},$$$$\frac{\Gamma(z+\alpha)}{\Gamma(z+\beta)}=\sum_{n=O}^\infty C_n(\alpha-\beta,\beta)z^{\alpha-\beta-n},$$ 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 $\sqrt{i}=(1+i)/\sqrt{2}$.

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 ) $$\frac{\Gamma(z+\alpha)}{\Gamma(z+\beta)}=\sum_{n=O}^\infty C_n(\alpha-\beta,\beta)x^{\alpha-\beta-n},$$ 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 only terms $n\gg m$ are relevant. 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 $\sqrt{i}=(1+i)/\sqrt{2}$.

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 $\sqrt{i}=(1+i)/\sqrt{2}$.

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created Mar 10, 2018 at 19:34

Zurab Silagadze

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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 ) $$\frac{\Gamma(z+\alpha)}{\Gamma(z+\beta)}=\sum_{n=O}^\infty C_n(\alpha-\beta,\beta)x^{\alpha-\beta-n},$$ 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 only terms $n\gg m$ are relevant. 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 $\sqrt{i}=(1+i)/\sqrt{2}$.