Watson's lemma explained

In mathematics, Watson's lemma, proved by G. N. Watson (1918, p. 133), has significant application within the theory on the asymptotic behavior of integrals.

Statement of the lemma

Let

0 < T \leq inf ty be fixed. Assume \varphi(t) = t ^{λg(t)} , where g(t) has an infinite number of derivatives in the neighborhood of t =0 , with g( 0 ) ≠ 0 , and λ > -1 .

Suppose, in addition, either that

|\varphi(t)| < Ke ^{bt} \forall t> 0 ,

where

K,b are independent of t , or that\int ^{T} _{0} |\varphi(t)| dt < inf ty.

Then, it is true that for all positive

x that\left|\int ^{T} _{0} e ^{-x} \varphi(t) dt\right| < inf ty

and that the following asymptotic equivalence holds:

\int ^{T} _{0} e ^{-x} \varphi(t) dt \sim \sum ^{inf ty} _{n =0} (g ^{(n)} ( 0 ) \Gamma( λ+ n +1 ))/(n! x ^{λ+n+1}) , (x> 0 , x → inf ty).

See, for instance, for the original proof or for a more recent development.

Proof

We will prove the version of Watson's lemma which assumes that

|\varphi(t)| has at most exponential growth as t \to inf ty . The basic idea behind the proof is that we will approximate g(t) by finitely many terms of its Taylor series. Since the derivatives of g are only assumed to exist in a neighborhood of the origin, we will essentially proceed by removing the tail of the integral, applying Taylor's theorem with remainder in the remaining small interval, then adding the tail back on in the end. At each step we will carefully estimate how much we are throwing away or adding on. This proof is a modification of the one found in .

Let

0 < T \leq inf ty and suppose that \varphi is a measurable function of the form \varphi(t) = t ^{λ} g(t) , where λ > -1 and g has an infinite number of continuous derivatives in the interval [ 0 ,\delta] for some 0 < \delta < T , and that |\varphi(t)| \leq Ke ^{bt} for all \delta \leq t \leq T , where the constants K and b are independent of t .

We can show that the integral is finite for

x large enough by writing( 1 ) \int ^{T} _{0} e ^{-xt} \varphi(t) dt = \int ^{\delta} _{0} e ^{-xt} \varphi(t) dt + \int ^{T} _{\delta} e ^{-xt} \varphi(t) dt and estimating each term.

For the first term we have

\left|\int ^{\delta} _{0} e ^{-xt} \varphi(t) dt\right| \leq \int ^{\delta} _{0} e ^{-xt} |\varphi(t)| dt \leq \int ^{\delta} _{0} |\varphi(t)| dt for x \geq 0 , where the last integral is finite by the assumptions that g is continuous on the interval [ 0 ,\delta] and that λ > -1 . For the second term we use the assumption that \varphi is exponentially bounded to see that, for x > b ,\begin{align} \left|\int ^{T} _{\delta} e ^{-xt} \varphi(t) dt\right| &\leq \int ^{T} _{\delta} e ^{-xt} |\varphi(t)| dt \\ &\leq K \int ^{T} _{\delta} e ^{(b-x)t} dt \\ &\leq K \int ^{inf ty} _{\delta} e ^{(b-x)t} dt \\ & = K (e ^{(b-x)\delta})/(x - b) . \end{align} The finiteness of the original integral then follows from applying the triangle inequality to ( 1 ) .

We can deduce from the above calculation that

( 2 ) \int ^{T} _{0} e ^{-xt} \varphi(t) dt = \int ^{\delta} _{0} e ^{-xt} \varphi(t) dt + O\left(x ^{-1} e ^{-\delta} \right) as x \to inf ty .

By appealing to Taylor's theorem with remainder we know that, for each integer

N \geq 0 ,g(t) = \sum ^{N} _{n =0} (g ^{(n)} ( 0 ))/(n!) t ^{n} + (g ^{(N+1)} (t ^{*)})/((N +1 )!) t ^{N+1} for 0 \leq t \leq \delta , where 0 \leq t ^{*} \leq t . Plugging this in to the first term in ( 2 ) we get\begin{align} ( 3 ) \int ^{\delta} _{0} e ^{-xt} \varphi(t) dt & = \int ^{\delta} _{0} e ^{-xt} t ^{λ} g(t) dt \\ & = \sum ^{N} _{n =0} (g ^{(n)} ( 0 ))/(n!) \int ^{\delta} _{0} t ^{λ} e ^{-xt} dt + (1)/((N +1 )!) \int ^{\delta} _{0} g ^{(N+1)} (t ^{*)} t ^{λ+N+1} e ^{-xt} dt. \end{align} To bound the term involving the remainder we use the assumption that g ^{(N+1)} is continuous on the interval [ 0 ,\delta] , and in particular it is bounded there. As such we see that\begin{align} \left|\int ^{\delta} _{0} g ^{(N+1)} (t ^{*)} t ^{λ+N+1} e ^{-xt} dt\right| &\leq \sup _{t} \left|g ^{(N+1)} (t)\right| \int ^{\delta} _{0} t ^{λ+N+1} e ^{-xt} dt \\ &< \sup _{t} \left|g ^{(N+1)} (t)\right| \int ^{inf ty} _{0} t ^{λ+N+1} e ^{-xt} dt \\ & = \sup _{t} \left|g ^{(N+1)} (t)\right| (\Gamma( λ+ N +2 ))/(x ^{λ+N+2}) . \end{align} Here we have used the fact that\int ^{inf ty} _{0} t ^{a} e ^{-xt} dt = (\Gamma(a +1 ))/(x ^{a+1}) if x > 0 and a > -1 , where \Gamma is the gamma function.

From the above calculation we see from

( 3 ) that( 4 ) \int ^{\delta} _{0} e ^{-xt} \varphi(t) dt = \sum ^{N} _{n =0} (g ^{(n)} ( 0 ))/(n!) \int ^{\delta} _{0} t ^{λ} e ^{-xt} dt + O\left(x ^{-λ-N-2} \right) as x \to inf ty .

We will now add the tails on to each integral in

( 4 ) . For each n we have\begin{align} \int ^{\delta} _{0} t ^{λ} e ^{-xt} dt & = \int ^{inf ty} _{0} t ^{λ} e ^{-xt} dt - \int ^{inf ty} _{\delta} t ^{λ} e ^{-xt} dt \\[ 5 pt] & = (\Gamma( λ+ n +1 ))/(x ^{λ+n+1}) - \int ^{inf ty} _{\delta} t ^{λ} e ^{-xt} dt, \end{align} and we will show that the remaining integrals are exponentially small. Indeed, if we make the change of variables t = s + \delta we get\begin{align} \int ^{inf ty} _{\delta} t ^{λ} e ^{-xt} dt & = \int ^{inf ty} _{0} (s + \delta) ^{λ} e ^{-x(s+\delta)} d s \\[ 5 pt] & = e ^{-\delta} \int ^{inf ty} _{0} (s + \delta) ^{λ} e ^{-xs} d s \\[ 5 pt] &\leq e ^{-\delta} \int ^{inf ty} _{0} (s + \delta) ^{λ} e ^{-s} d s \end{align} for x \geq 1 , so that\int ^{\delta} _{0} t ^{λ} e ^{-xt} dt = (\Gamma( λ+ n +1 ))/(x ^{λ+n+1}) + O\left(e ^{-\delta} \right) as x \to inf ty.

If we substitute this last result into

( 4 ) we find that\begin{align} \int ^{\delta} _{0} e ^{-xt} \varphi(t) dt & = \sum ^{N} _{n =0} (g ^{(n)} ( 0 ) \Gamma( λ+ n +1 ))/(n! x ^{λ+n+1}) + O\left(e ^{-\delta} \right) + O\left(x ^{-λ-N-2} \right) \\ & = \sum ^{N} _{n =0} (g ^{(n)} ( 0 ) \Gamma( λ+ n +1 ))/(n! x ^{λ+n+1}) + O\left(x ^{-λ-N-2} \right) \end{align} as x \to inf ty . Finally, substituting this into ( 2 ) we conclude that\begin{align} \int ^{T} _{0} e ^{-xt} \varphi(t) dt & = \sum ^{N} _{n =0} (g ^{(n)} ( 0 ) \Gamma( λ+ n +1 ))/(n! x ^{λ+n+1}) + O\left(x ^{-λ-N-2} \right) + O\left(x ^{-1} e ^{-\delta} \right) \\ & = \sum ^{N} _{n =0} (g ^{(n)} ( 0 ) \Gamma( λ+ n +1 ))/(n! x ^{λ+n+1}) + O\left(x ^{-λ-N-2} \right) \end{align} as x \to inf ty .

Since this last expression is true for each integer

N \geq 0 we have thus shown that\int ^{T} _{0} e ^{-xt} \varphi(t) dt \sim \sum ^{inf ty} _{n =0} (g ^{(n)} ( 0 ) \Gamma( λ+ n +1 ))/(n! x ^{λ+n+1}) as x \to inf ty , where the infinite series is interpreted as an asymptotic expansion of the integral in question.

Example

When

0 < a < b , the confluent hypergeometric function of the first kind has the integral representation{} _{1F} _{1(a,b,x)} = (\Gamma(b))/(\Gamma(a) \Gamma(b - a)) \int ^{1} _{0} e ^{xt} t ^{a-1} ( 1- t) ^{b-a-1} dt, where \Gamma is the gamma function. The change of variables t = 1- s puts this into the form{} _{1F} _{1(a,b,x)} = (\Gamma(b))/(\Gamma(a) \Gamma(b - a)) e ^{1} _{0} e ^{-xs} ( 1- s) ^{a-1} s ^{b-a-1} ds, which is now amenable to the use of Watson's lemma. Taking λ = b - a -1 and g(s) = ( 1- s) ^{a-1} , Watson's lemma tells us that\int ^{1} _{0} e ^{-xs} ( 1- s) ^{a-1} s ^{b-a-1} ds \sim \Gamma(b - a) x ^{a-b} as x \to inf ty with x > 0 , which allows us to conclude that{} _{1F} _{1(a,b,x)} \sim (\Gamma(b))/(\Gamma(a)) x ^{a-b} e ^{x} as x \to inf ty with x > 0 .

References

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• Ablowitz, M. J., Fokas, A. S. (2003). Complex variables: introduction and applications. Cambridge University Press.