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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 $0<T\le \mathrm{\infty }$ be fixed. Assume $\varphi (t)=t^{\lambda }g(t)$, where $g(t)$ has an infinite number of derivatives in the neighborhood of $t=0$, with $g(0)\ne 0$, and $\lambda >-1$.

Suppose, in addition, either that

$|\varphi (t)|<K{e}^{bt}\mathrm{\forall }t>0,$

where $K,b$ are independent of $t$, or that

$\underset{0}{\overset{T}{\int }}|\varphi (t)|dt<\mathrm{\infty }.$

Then, it is true that for all positive $x$ that

$|\underset{0}{\overset{T}{\int }}{e}^{-xt}\varphi (t)dt|<\mathrm{\infty }$

and that the following asymptotic equivalence holds:

$\underset{0}{\overset{T}{\int }}{e}^{-xt}\varphi (t)dt\sim \sum _{n=0}^{\mathrm{\infty }}\frac{g^{(n)}(0)\mathrm{\Gamma }(\lambda +n+1)}{n!x^{\lambda +n+1}},(x>0,x\to \mathrm{\infty }).$

See, for instance, Template:Harvtxt for the original proof or Template:Harvtxt 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 $t\to \mathrm{\infty }$. 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 Template:Harvtxt.

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

We can show that the integral is finite for $x$ large enough by writing

$(1)\underset{0}{\overset{T}{\int }}{e}^{-xt}\varphi (t)dt=\underset{0}{\overset{\delta }{\int }}{e}^{-xt}\varphi (t)dt+\underset{\delta }{\overset{T}{\int }}{e}^{-xt}\varphi (t)dt$

and estimating each term.

For the first term we have

$|\underset{0}{\overset{\delta }{\int }}{e}^{-xt}\varphi (t)dt|\le \underset{0}{\overset{\delta }{\int }}{e}^{-xt}|\varphi (t)|dt\le \underset{0}{\overset{\delta }{\int }}|\varphi (t)|dt$

for $x\ge 0$, where the last integral is finite by the assumptions that $g$ is continuous on the interval $[0,\delta ]$ and that $\lambda >-1$. For the second term we use the assumption that $\varphi$ is exponentially bounded to see that, for $x>b$,

$\begin{array}{rl}|\underset{\delta }{\overset{T}{\int }}{e}^{-xt}\varphi (t)dt|& \le \underset{\delta }{\overset{T}{\int }}{e}^{-xt}|\varphi (t)|dt\\ & \le K\underset{\delta }{\overset{T}{\int }}{e}^{(b-x)t}dt\\ & \le K\underset{\delta }{\overset{\mathrm{\infty }}{\int }}{e}^{(b-x)t}dt\\ & =K\frac{e^{(b-x)\delta }}{x-b}.\end{array}$

The finiteness of the original integral then follows from applying the triangle inequality to $(1)$.

We can deduce from the above calculation that

$(2)\underset{0}{\overset{T}{\int }}{e}^{-xt}\varphi (t)dt=\underset{0}{\overset{\delta }{\int }}{e}^{-xt}\varphi (t)dt+O\left(x^{-1}{e}^{-\delta x}\right)$

as $x\to \mathrm{\infty }$.

By appealing to Taylor's theorem with remainder we know that, for each integer $N\ge 0$,

$g(t)=\sum _{n=0}^N\frac{g^{(n)}(0)}{n!}{t}^n+\frac{g^{(N+1)}(t^{*})}{(N+1)!}{t}^{N+1}$

for $0\le t\le \delta$, where $0\le t^{*}\le t$. Plugging this in to the first term in $(2)$ we get

$\begin{array}{rl}(3)\underset{0}{\overset{\delta }{\int }}{e}^{-xt}\varphi (t)dt& =\underset{0}{\overset{\delta }{\int }}{e}^{-xt}{t}^{\lambda }g(t)dt\\ & =\sum _{n=0}^N\frac{g^{(n)}(0)}{n!}\underset{0}{\overset{\delta }{\int }}{t}^{\lambda +n}{e}^{-xt}dt+\frac{1}{(N+1)!}\underset{0}{\overset{\delta }{\int }}{g}^{(N+1)}(t^{*})t^{\lambda +N+1}{e}^{-xt}dt.\end{array}$

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

$\begin{array}{rl}|\underset{0}{\overset{\delta }{\int }}{g}^{(N+1)}(t^{*})t^{\lambda +N+1}{e}^{-xt}dt|& \le {\sup }_{t\in [0,\delta ]}|g^{(N+1)}(t)|\underset{0}{\overset{\delta }{\int }}{t}^{\lambda +N+1}{e}^{-xt}dt\\ & <{\sup }_{t\in [0,\delta ]}|g^{(N+1)}(t)|\underset{0}{\overset{\mathrm{\infty }}{\int }}{t}^{\lambda +N+1}{e}^{-xt}dt\\ & ={\sup }_{t\in [0,\delta ]}|g^{(N+1)}(t)|\frac{\mathrm{\Gamma }(\lambda +N+2)}{x^{\lambda +N+2}}.\end{array}$

Here we have used the fact that

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{t}^a{e}^{-xt}dt=\frac{\mathrm{\Gamma }(a+1)}{x^{a+1}}$

if $x>0$ and $a>-1$, where $\mathrm{\Gamma }$ is the gamma function.

From the above calculation we see from $(3)$ that

$(4)\underset{0}{\overset{\delta }{\int }}{e}^{-xt}\varphi (t)dt=\sum _{n=0}^N\frac{g^{(n)}(0)}{n!}\underset{0}{\overset{\delta }{\int }}{t}^{\lambda +n}{e}^{-xt}dt+O\left(x^{-\lambda -N-2}\right)$

as $x\to \mathrm{\infty }$.

We will now add the tails on to each integral in $(4)$. For each $n$ we have

$\begin{array}{rl}\underset{0}{\overset{\delta }{\int }}{t}^{\lambda +n}{e}^{-xt}dt& =\underset{0}{\overset{\mathrm{\infty }}{\int }}{t}^{\lambda +n}{e}^{-xt}dt-\underset{\delta }{\overset{\mathrm{\infty }}{\int }}{t}^{\lambda +n}{e}^{-xt}dt\\ & =\frac{\mathrm{\Gamma }(\lambda +n+1)}{x^{\lambda +n+1}}-\underset{\delta }{\overset{\mathrm{\infty }}{\int }}{t}^{\lambda +n}{e}^{-xt}dt,\end{array}$

and we will show that the remaining integrals are exponentially small. Indeed, if we make the change of variables $t=s+\delta$ we get

$\begin{array}{rl}\underset{\delta }{\overset{\mathrm{\infty }}{\int }}{t}^{\lambda +n}{e}^{-xt}dt& =\underset{0}{\overset{\mathrm{\infty }}{\int }}(s+\delta {)}^{\lambda +n}{e}^{-x(s+\delta )}ds\\ & =e^{-\delta x}\underset{0}{\overset{\mathrm{\infty }}{\int }}(s+\delta {)}^{\lambda +n}{e}^{-xs}ds\\ & \le e^{-\delta x}\underset{0}{\overset{\mathrm{\infty }}{\int }}(s+\delta {)}^{\lambda +n}{e}^{-s}ds\end{array}$

for $x\ge 1$, so that

$\underset{0}{\overset{\delta }{\int }}{t}^{\lambda +n}{e}^{-xt}dt=\frac{\mathrm{\Gamma }(\lambda +n+1)}{x^{\lambda +n+1}}+O\left(e^{-\delta x}\right)$

as $x\to \mathrm{\infty }$.

If we substitute this last result into $(4)$ we find that

$\begin{array}{rl}\underset{0}{\overset{\delta }{\int }}{e}^{-xt}\varphi (t)dt& =\sum _{n=0}^N\frac{g^{(n)}(0)\mathrm{\Gamma }(\lambda +n+1)}{n!x^{\lambda +n+1}}+O\left(e^{-\delta x}\right)+O\left(x^{-\lambda -N-2}\right)\\ & =\sum _{n=0}^N\frac{g^{(n)}(0)\mathrm{\Gamma }(\lambda +n+1)}{n!x^{\lambda +n+1}}+O\left(x^{-\lambda -N-2}\right)\end{array}$

as $x\to \mathrm{\infty }$. Finally, substituting this into $(2)$ we conclude that

$\begin{array}{rl}\underset{0}{\overset{T}{\int }}{e}^{-xt}\varphi (t)dt& =\sum _{n=0}^N\frac{g^{(n)}(0)\mathrm{\Gamma }(\lambda +n+1)}{n!x^{\lambda +n+1}}+O\left(x^{-\lambda -N-2}\right)+O\left(x^{-1}{e}^{-\delta x}\right)\\ & =\sum _{n=0}^N\frac{g^{(n)}(0)\mathrm{\Gamma }(\lambda +n+1)}{n!x^{\lambda +n+1}}+O\left(x^{-\lambda -N-2}\right)\end{array}$

as $x\to \mathrm{\infty }$.

Since this last expression is true for each integer $N\ge 0$ we have thus shown that

$\underset{0}{\overset{T}{\int }}{e}^{-xt}\varphi (t)dt\sim \sum _{n=0}^{\mathrm{\infty }}\frac{g^{(n)}(0)\mathrm{\Gamma }(\lambda +n+1)}{n!x^{\lambda +n+1}}$

as $x\to \mathrm{\infty }$, 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

${}_1{F}_1(a,b,x)=\frac{\mathrm{\Gamma }(b)}{\mathrm{\Gamma }(a)\mathrm{\Gamma }(b-a)}\underset{0}{\overset{1}{\int }}{e}^{xt}{t}^{a-1}(1-t{)}^{b-a-1}dt,$

where $\mathrm{\Gamma }$ is the gamma function. The change of variables $t=1-s$ puts this into the form

${}_1{F}_1(a,b,x)=\frac{\mathrm{\Gamma }(b)}{\mathrm{\Gamma }(a)\mathrm{\Gamma }(b-a)}{e}^x\underset{0}{\overset{1}{\int }}{e}^{-xs}(1-s{)}^{a-1}{s}^{b-a-1}ds,$

which is now amenable to the use of Watson's lemma. Taking $\lambda =b-a-1$ and $g(s)=(1-s{)}^{a-1}$, Watson's lemma tells us that

$\underset{0}{\overset{1}{\int }}{e}^{-xs}(1-s{)}^{a-1}{s}^{b-a-1}ds\sim \mathrm{\Gamma }(b-a)x^{a-b}\text{as }x\to \mathrm{\infty }\text{ with }x>0,$

which allows us to conclude that

${}_1{F}_1(a,b,x)\sim \frac{\mathrm{\Gamma }(b)}{\mathrm{\Gamma }(a)}{x}^{a-b}{e}^x\text{as }x\to \mathrm{\infty }\text{ with }x>0.$

References

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• Many property agents need to declare for the PIC grant in Singapore. However, not all of them know find out how to do the correct process for getting this PIC scheme from the IRAS. There are a number of steps that you need to do before your software can be approved.
  
  Naturally, you will have to pay a safety deposit and that is usually one month rent for annually of the settlement. That is the place your good religion deposit will likely be taken into account and will kind part or all of your security deposit. Anticipate to have a proportionate amount deducted out of your deposit if something is discovered to be damaged if you move out. It's best to you'll want to test the inventory drawn up by the owner, which can detail all objects in the property and their condition. If you happen to fail to notice any harm not already mentioned within the inventory before transferring in, you danger having to pay for it yourself.
  
  In case you are in search of an actual estate or Singapore property agent on-line, you simply should belief your intuition. It's because you do not know which agent is nice and which agent will not be. Carry out research on several brokers by looking out the internet. As soon as if you end up positive that a selected agent is dependable and reliable, you can choose to utilize his partnerise in finding you a home in Singapore. Most of the time, a property agent is taken into account to be good if he or she locations the contact data on his website. This may mean that the agent does not mind you calling them and asking them any questions relating to new properties in singapore in Singapore. After chatting with them you too can see them in their office after taking an appointment.
  
  Have handed an trade examination i.e Widespread Examination for House Brokers (CEHA) or Actual Property Agency (REA) examination, or equal; Exclusive brokers are extra keen to share listing information thus making certain the widest doable coverage inside the real estate community via Multiple Listings and Networking. Accepting a severe provide is simpler since your agent is totally conscious of all advertising activity related with your property. This reduces your having to check with a number of agents for some other offers. Price control is easily achieved. Paint work in good restore-discuss with your Property Marketing consultant if main works are still to be done. Softening in residential property prices proceed, led by 2.8 per cent decline within the index for Remainder of Central Region
  
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  15.6.1 As the agent is an intermediary, generally, as soon as the principal and third party are introduced right into a contractual relationship, the agent drops out of the image, subject to any problems with remuneration or indemnification that he could have against the principal, and extra exceptionally, against the third occasion. Generally, agents are entitled to be indemnified for all liabilities reasonably incurred within the execution of the brokers´ authority.
  
  To achieve the very best outcomes, you must be always updated on market situations, including past transaction information and reliable projections. You could review and examine comparable homes that are currently available in the market, especially these which have been sold or not bought up to now six months. You'll be able to see a pattern of such report by clicking here It's essential to defend yourself in opposition to unscrupulous patrons. They are often very skilled in using highly unethical and manipulative techniques to try and lure you into a lure. That you must also protect your self, your loved ones, and personal belongings as you'll be serving many strangers in your home. Sign a listing itemizing of all of the objects provided by the proprietor, together with their situation. HSR Prime Recruiter 2010.