TeachingSpecial Functions – Asymptotics – Watson's lemma
We will prove Watson's lemma:
Lemma: Let \(f\) be a complex valued function of a real variable \(t\) such that
- \(f\) is continuous on \((0,\infty)\),
- \(f(t)\sim\displaystyle\sum_{n=0}^{\infty}a_nt^{\lambda_n-1}\quad\text{for}\quad t\downarrow 0\) with \(0<\lambda_0<\lambda_1<\lambda_2<\cdots\) and
- for some fixed \(c>0\): \(f(t)=\mathcal{O}\left(e^{ct}\right)\) for \(t\rightarrow\infty\).
Then we have
\[F(z)=\int_0^{\infty}e^{-zt}f(t)\,dt\sim\sum_{n=0}^{\infty}a_n\frac{\Gamma(\lambda_n)}{z^{\lambda_n}} \quad\text{for}\quad|z|\to\infty\quad\text{and}\quad\left|\arg(z)\right|\leq\frac{\pi}{2}-\delta<\frac{\pi}{2}\]for some \(\delta\) such that \(0<\delta<\pi/2\).
Proof: From the theory of Laplace transforms it is well known that
\[F(z)=\int_0^{\infty}e^{-zt}f(t)\,dt,\]the Laplace transform of \(f\), exists for \(\text{Re}(z)>c\) if \(f\) satisfies the three conditions mentioned in the lemma. That is: the integral converges for \(\text{Re}(z)>c\). Note that the second condition implies that
\[\left|f(t)-\sum_{n=0}^{N-1}a_nt^{\lambda_n-1}\right|\le M\,t^{\lambda_N-1}\quad\text{for}\quad t\downarrow 0,\]where \(M>0\) is some constant. Together with the third condition this implies that
\[\left|f(t)-\sum_{n=0}^{N-1}a_nt^{\lambda_n-1}\right|\le K\,e^{ct}t^{\lambda_N-1}\quad\text{for}\quad t>0,\]where \(K>0\) is some constant. Hence we have
\[\left|\int_0^{\infty}e^{-zt}f(t)\,dt-\sum_{n=0}^{N-1}a_n\int_0^{\infty}e^{-zt}t^{\lambda_n-1}\,dt\right| \leq K\int_0^{\infty}e^{-(\text{Re}(z)-c)t}t^{\lambda_N-1}\,dt.\]Note that we have for \(\text{Re}(z)>0\)
\[\int_0^{\infty}e^{-zt}t^{\lambda_n-1}\,dt=\frac{1}{z^{\lambda_n}}\int_0^{\infty}e^{-\tau}\tau^{\lambda_n-1}\,d\tau =\frac{\Gamma(\lambda_n)}{z^{\lambda_n}}.\]Hence we have
\[\left|F(z)-\sum_{n=0}^{N-1}a_n\frac{\Gamma(\lambda_n)}{z^{\lambda_n}}\right| \leq K\,\frac{\Gamma(\lambda_N)}{\left(\text{Re}(z)-c\right)^{\lambda_N}} =K\,\frac{\Gamma(\lambda_N)}{|z|^{\lambda_N}}\left(\frac{|z|}{\text{Re}(z)-c}\right)^{\lambda_N}.\]Since \(|\arg(z)|\leq\frac{\pi}{2}-\delta<\frac{\pi}{2}\), we have \(\text{Re}(z)\geq |z|\sin\delta\) which implies that \(\text{Re}(z)-c\geq\frac{1}{2}|z|\sin\delta\) for \(|z|\) large enough. This implies that we have
\[F(z)-\sum_{n=0}^{N-1}a_n\frac{\Gamma(\lambda_n)}{z^{\lambda_n}}=\mathcal{O}\left(z^{-\lambda_N}\right),\]which proves Watson's lemma.
Some examples of the application of Watson's lemma:
1) Consider the function \(f(t)=1/(1+t)\). Then we have: \(f\) is continuous on \((0,\infty)\) and
\[f(t)=\frac{1}{1+t}=\sum_{n=0}^{\infty}(-1)^nt^n,\quad |t|<1.\]Now Watson's lemma implies that
\[F(z)=\int_0^{\infty}e^{-zt}f(t)\,dt=\int_0^{\infty}\frac{e^{-zt}}{1+t}\,dt\sim \sum_{n=0}^{\infty}(-1)^n\frac{\Gamma(n+1)}{z^{n+1}}=\sum_{n=0}^{\infty}(-1)^n\frac{n!}{z^{n+1}}\quad\text{for}\quad |z|\to\infty\quad\text{and}\quad\left|\arg(z)\right|\leq\frac{\pi}{2}-\delta<\frac{\pi}{2}.\]2) Consider the function \(f(t)=1/\sqrt{1+t^2}\). Then we have: \(f\) is continuous on \((0,\infty)\) and
\[f(t)=\frac{1}{\sqrt{1+t^2}}=(1+t^2)^{-\frac{1}{2}} =\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n}{n!}t^{2n},\quad |t|<1.\]Now Watson's lemma implies that
\[F(z)=\int_0^{\infty}e^{-zt}f(t)\,dt=\int_0^{\infty}\frac{e^{-zt}}{\sqrt{1+t^2}}\,dt\sim \sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n}{n!}\cdot\frac{(2n)!}{z^{2n+1}}\quad\text{for}\quad|z|\to\infty \quad\text{and}\quad\left|\arg(z)\right|\leq\frac{\pi}{2}-\delta<\frac{\pi}{2}.\]Since we have
\[\left(\frac{1}{2}\right)_n=\frac{1}{2}\cdot\frac{3}{2}\cdots\left(\frac{1}{2}+n-1\right) =\frac{1\cdot3\cdots(2n-1)}{2^n}\cdot\frac{2\cdot4\cdots(2n)}{2^n\,n!}=\frac{(2n)!}{2^{2n}\,n!},\]this can also be written as
\[\int_0^{\infty}\frac{e^{-zt}}{\sqrt{1+t^2}}\,dt\sim \sum_{n=0}^{\infty}(-1)^n\left(\frac{(2n)!}{2^n\,n!}\right)^2\frac{1}{z^{2n+1}}\quad\text{for}\quad|z|\to\infty \quad\text{and}\quad\left|\arg(z)\right|\leq\frac{\pi}{2}-\delta<\frac{\pi}{2}.\]3) Consider the function \(f(t)=\ln(1+t^2)\). Then we have: \(f\) is continuous on \((0,\infty)\) and
\[f(t)=\ln(1+t^2)=\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}t^{2n+2},\quad |t|<1.\]Now Watson's lemma implies that
\[F(z)=\int_0^{\infty}e^{-zt}f(t)\,dt=\int_0^{\infty}e^{-zt}\ln(1+t^2)\,dt\sim \sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}\cdot\frac{(2n+2)!}{z^{2n+3}}\quad\text{for}\quad|z|\to\infty\quad\text{and}\quad \left|\arg(z)\right|\leq\frac{\pi}{2}-\delta<\frac{\pi}{2}.\]Note that this can also be written as
\[\int_0^{\infty}e^{-zt}\ln(1+t^2)\,dt\sim2\sum_{n=0}^{\infty}(-1)^n\frac{(2n+1)!}{z^{2n+3}}\quad\text{for}\quad |z|\to\infty\quad\text{and}\quad\left|\arg(z)\right|\leq\frac{\pi}{2}-\delta<\frac{\pi}{2}.\]4) For \(\text{Re}(c)>\text{Re}(a)>0\) the confluent hypergeometric function can be written as
\[{}_1F_1\left(\genfrac{}{}{0pt}{}{a}{c}\,;\,z\right)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int_0^1e^{z\tau}\tau^{a-1}(1-\tau)^{c-a-1}\,d\tau.\]Now we use the substitution \(\tau=1-t\) to find that
\[{}_1F_1\left(\genfrac{}{}{0pt}{}{a}{c}\,;\,z\right)=\frac{\Gamma(c)\,e^z}{\Gamma(a)\Gamma(c-a)} \int_0^1e^{-zt}(1-t)^{a-1}t^{c-a-1}\,dt=\int_0^{\infty}e^{-zt}f(t)\,dt,\]where
\[f(t)=\frac{\Gamma(c)\,e^z}{\Gamma(a)\Gamma(c-a)}(1-t)^{a-1}t^{c-a-1}\quad\text{for}\quad0 < t < 1\]and \(f(t)=0\) for \(t\geq 1\). Note that we have
\[f(t)=\frac{\Gamma(c)\,e^z}{\Gamma(a)\Gamma(c-a)}\sum_{n=0}^{\infty} (-1)^n\binom{a-1}{n}t^{n+c-a-1}\quad\text{for}\quad t\downarrow 0.\]Now Watson's lemma implies that
\[{}_1F_1\left(\genfrac{}{}{0pt}{}{a}{c}\,;\,z\right)\sim\frac{\Gamma(c)\,e^z}{\Gamma(a)\Gamma(c-a)}\sum_{n=0}^{\infty} (-1)^n\binom{a-1}{n}\frac{\Gamma(n+c-a)}{z^{n+c-a}}\quad\text{for}\quad|z|\to\infty\quad\text{and}\quad \left|\arg(z)\right|\leq\frac{\pi}{2}-\delta<\frac{\pi}{2}.\]Since we have
\[\frac{\Gamma(n+c-a)}{\Gamma(c-a)}=(c-a)_n\quad\text{and}\quad (-1)^n\binom{a-1}{n}=\frac{(1-a)_n}{n!},\]this can also be written as
\[{}_1F_1\left(\genfrac{}{}{0pt}{}{a}{c}\,;\,z\right)\sim\frac{\Gamma(c)}{\Gamma(a)}\,e^z\,z^{a-c}\sum_{n=0}^{\infty}\frac{(1-a)_n(c-a)_n}{n!}\frac{1}{z^n} =\frac{\Gamma(c)}{\Gamma(a)}\,e^z\,z^{a-c}{}_2F_0\left(\genfrac{}{}{0pt}{}{1-a,\,c-a}{-}\,;\,\frac{1}{z}\right)\]for \(|z|\to\infty\) and \(\left|\arg(z)\right|\leq\frac{\pi}{2}-\delta<\frac{\pi}{2}\).
Finally, we derive two asymptotic expansions for the Bessel function of the first kind.
For \(\text{Re}(\nu)>-1/2\) the Bessel function of the first kind of order \(\nu\) can be written as
\[J_{\nu}(z)=\frac{1}{\Gamma(\nu+\frac{1}{2})\sqrt{\pi}}\left(\frac{z}{2}\right)^{\nu} \int_{-1}^1e^{izt}(1-t^2)^{\nu-\frac{1}{2}}\,dt.\]Now we use the substitution \(t=2\tau-1\) to obtain for \(\text{Re}(\nu)>-1/2\)
\[J_{\nu}(z)=\frac{2^{2\nu}e^{-iz}}{\Gamma(\nu+\frac{1}{2})\sqrt{\pi}}\left(\frac{z}{2}\right)^{\nu} \int_0^1e^{2iz\tau}\tau^{\nu-\frac{1}{2}}(1-\tau)^{\nu-\frac{1}{2}}\,d\tau.\]For \(\text{Re}(c)>\text{Re}(a)>0\) we have
\[{}_1F_1\left(\genfrac{}{}{0pt}{}{a}{c}\,;\,z\right)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int_0^1e^{zt}t^{a-1}(1-t)^{c-a-1}\,dt,\]which implies that for \(\text{Re}(\nu)>-1/2\) we have
\[J_{\nu}(z)=\frac{2^{2\nu}e^{-iz}}{\Gamma(\nu+\frac{1}{2})\sqrt{\pi}}\left(\frac{z}{2}\right)^{\nu} \frac{\Gamma(\nu+\frac{1}{2})\Gamma(\nu+\frac{1}{2})}{\Gamma(2\nu+1)} {}_1F_1\left(\genfrac{}{}{0pt}{}{\textstyle\nu+\frac{1}{2}}{2\nu+1}\,;\,2iz\right).\]Now we use the asymptotic expansion for the confluent hypergeometric function to obtain
\[J_{\nu}(z)\sim\frac{2^{2\nu}e^{-iz}}{\Gamma(\nu+\frac{1}{2})\sqrt{\pi}} \left(\frac{z}{2}\right)^{\nu}e^{2iz}\sum_{n=0}^{\infty}(-1)^n \binom{\nu-\frac{1}{2}}{n}\frac{\Gamma(n+\nu+\frac{1}{2})}{(2iz)^{n+\nu+\frac{1}{2}}}\]for \(|z|\to\infty\) and \(\left|\arg(2iz)\right|\leq\frac{\pi}{2}-\delta<\frac{\pi}{2}\). Now we use
\[\frac{1}{\Gamma(\nu+\frac{1}{2})}\binom{\nu-\frac{1}{2}}{n} =\frac{1}{\Gamma(\nu+\frac{1}{2})}\cdot\frac{\Gamma(\nu+\frac{1}{2})}{n!\,\Gamma(\nu-n+\frac{1}{2})} =\frac{1}{\Gamma(\nu-n+\frac{1}{2})\,n!}\]and the fact that \(i=e^{\pi i/2}\) to obtain
\[J_{\nu}(z)\sim\frac{1}{\sqrt{2\pi z}}\,e^{i\left(z-\frac{\pi\nu}{2}-\frac{\pi}{4}\right)} \sum_{n=0}^{\infty}\frac{\Gamma(\nu+n+\frac{1}{2})}{\Gamma(\nu-n+\frac{1}{2})\,n!}\left(\frac{i}{2z}\right)^n\]for \(|z|\to\infty\) and \(\left|\arg(iz)\right|\leq\frac{\pi}{2}-\delta<\frac{\pi}{2}\) or equivalently \(-\pi<-\pi+\delta\leq\arg(z)\leq-\delta<0\).
In the same way the substitution \(t=1-2\tau\) leads to
\[J_{\nu}(z)\sim\frac{2^{2\nu}e^{iz}}{\Gamma(\nu+\frac{1}{2})\sqrt{\pi}} \left(\frac{z}{2}\right)^{\nu}e^{-2iz}\sum_{n=0}^{\infty}(-1)^n \binom{\nu-\frac{1}{2}}{n}\frac{\Gamma(n+\nu+\frac{1}{2})}{(-2iz)^{n+\nu+\frac{1}{2}}}\]for \(|z|\to\infty\) and \(\left|\arg(-2iz)\right|\leq\frac{\pi}{2}-\delta<\frac{\pi}{2}\). In that case we use the fact that \(-i=e^{-\pi/2}\) to obtain
\[J_{\nu}(z)\sim\frac{1}{\sqrt{2\pi z}}\,e^{-i\left(z-\frac{\pi\nu}{2}-\frac{\pi}{4}\right)} \sum_{n=0}^{\infty}\frac{\Gamma(\nu+n+\frac{1}{2})}{\Gamma(\nu-n+\frac{1}{2})\,n!}\left(\frac{1}{2iz}\right)^n\]for \(|z|\to\infty\) and \(\left|\arg(-iz)\right|\leq\frac{\pi}{2}-\delta<\frac{\pi}{2}\) or equivalently \(0<\delta\leq\arg(z)\le\pi-\delta<\pi\).
Last modified on October 2, 2021
