Homework 08
1 Asymptotic series for \(\log (x-2)\)
Find an asymptotic series for
\[ f(x)= \log (x-2) \]
as \(x \rightarrow \infty\) in terms of \(\log x\) and inverse powers of \(x\). (Hint: split the log and use Taylor series on one part.)
\[ f(x) = \log(x - 2) = \log x + \log(1 - \frac{2}{x}) \]
The Taylor series expansion of \(\log(1 - y)\) around \(y = 0\) is:
\[ \log(1 - y) = -y - \frac{y^2}{2} - \frac{y^3}{3} - \cdots \]
By substituting \(y = \frac{2}{x}\) into this series, the asymptotic series for \(f(x)\) as \(x \to \infty\) is:
\[ f(x) = \log x - \frac{2}{x} - \frac{2^2}{2x^2} - \frac{2^3}{3x^3} - \cdots \]
2 Watson’s lemma and \(\int_0^1 \frac{e^{-s x}}{1+x^2} d x\)
Use Watson’s lemma to find the \(s \rightarrow \infty\) asymptotic series for
\[ I(s)=\int_0^1 \frac{e^{-s x}}{1+x^2} d x \]
To use Watson’s Lemma, we need to express \(\frac{1}{1+x^2}\) as a power series at \(x = 0\). The Taylor series of \(\frac{1}{1+x^2}\) around \(x = 0\) is:
\[ \frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + \cdots \]
which is valid for \(|x| < 1\). Substituting this into \(I(s)\) gives:
\[ I(s) = \int_0^1 e^{-s x} (1 - x^2 + x^4 - x^6 + \cdots) \, dx \]
Now, we can integrate term by term:
For the first term:
\[ \int_0^1 e^{-s x} \, dx = \frac{1}{s} (1 - e^{-s}) \]
As \(s \rightarrow \infty\), \(e^{-s}\) approaches 0 faster than \(1/s\), so this term becomes \(\frac{1}{s}\).
For the second term:
\[ -\int_0^1 x^2 e^{-s x} \, dx \]
Applying integration by parts or a similar method, we find this term is \(O\left(\frac{1}{s^3}\right)\) as \(s \rightarrow \infty\).
Similarly, each subsequent term will contribute higher order terms in \(\frac{1}{s}\).
Hence, the asymptotic expansion of \(I(s)\) as \(s \rightarrow \infty\) is:
\[ I(s) \sim \frac{1}{s} - \frac{1}{s^3} + \cdots \]
3 Watson’s lemma and \(\int_0^{\infty} \sin (\sqrt{x}) e^{-s x^2} d x\)
Use Watson’s lemma to find the \(s \rightarrow \infty\) asymptotic series for
\[ I(s)=\int_0^{\infty} \sin (\sqrt{x}) e^{-s x^2} d x \]
The integral \(I(s)\) does not directly fit this form of Watson’s lemma due to the \(e^{-s x^2}\) term. So first, we perform a change of variable to transform the integral into a more suitable form for Watson’s Lemma. Let \(u = x^2\), then \(du = 2x \, dx\) or \(dx = \frac{du}{2\sqrt{u}}\). The integral becomes:
\[ I(s) = \int_0^{\infty} \sin(u^{1/4}) e^{-s u} \frac{du}{2\sqrt{u}} \]
Expand \(\sin(u^{1/4})\) in a Taylor series about \(u = 0\):
\[ \sin(u^{1/4}) = u^{1/4} - \frac{u^{3/4}}{3!} + \frac{u^{5/4}}{5!} - \cdots \]
Substituting this series into the integral, we get:
\[ I(s) = \int_0^{\infty} \left(u^{1/4} - \frac{u^{3/4}}{3!} + \cdots \right) e^{-s u} \frac{du}{2\sqrt{u}} = \frac{\Gamma(3/4)}{2 s^{3/4}} - \frac{\Gamma(5/4)}{12 s^{5/4}} + \cdots \]
4 Asymptotic series for the error function
Use integration by parts to give the \(x \rightarrow \infty\) asymptotic series for the error function
\[ \operatorname{erf}(x)=\frac{2}{\sqrt{\pi}} \int_x^{\infty} e^{-t^2} d t . \]
We can apply successive integration by parts to \(\operatorname{erfc}(\lambda)\), in which the first step gives
\[ \begin{aligned} \operatorname{erf}(x) & =\frac{2}{\sqrt{\pi}} \int_x^{\infty} \exp \left(-t^2\right) d t \\ & =\frac{2}{\sqrt{\pi}} \int_x^{\infty} \frac{-2 t \exp \left(-t^2\right)}{-2 t} d t,\left[u=\frac{1}{-2 t}, d v=-2 t \exp \left(-t^2\right) d t\right] \\ & =\frac{2}{\sqrt{\pi}}\left[\frac{\exp \left(-t^2\right)}{-2 t}-\int \frac{\exp \left(-t^2\right)}{2 t^2} d t\right]_x^{\infty} \\ & =\frac{2}{\sqrt{\pi}}\left[\frac{\exp \left(-x^2\right)}{2 x}-\int_x^{\infty} \frac{\exp \left(-t^2\right)}{2 t^2} d t\right] \\ & =\frac{2 e^{-x^2}}{\sqrt{\pi}}\left(\frac{1}{2 x}\right)-\frac{2}{\sqrt{\pi}} \int_x^{\infty} \frac{\exp \left(-t^2\right)}{2 t^2} d t \end{aligned} \]
We have obtained the first term. And applying integration by parts to the new integral gives
\[ \begin{aligned} \int_x^{\infty} \frac{\exp \left(-t^2\right)}{2 t^2} d t & =\frac{1}{2} \int_x^{\infty} \frac{-2 t \exp \left(-t^2\right)}{-2 t^3} d t\left[u=-\frac{1}{2 t^3}, d v=-2 t \exp \left(-t^2\right) d t\right] \\ & =\frac{1}{2}\left[\frac{\exp \left(-t^2\right)}{-2 t^3}-\int \frac{3 \exp \left(-t^2\right)}{2 t^4} d t\right]_x^{\infty} \\ & =\frac{1}{2}\left[\frac{\exp \left(-x^2\right)}{2 x^3}-\int_x^{\infty} \frac{3 \exp \left(-t^2\right)}{2 t^4} d t\right] \end{aligned} \]
Thus we have
\[ \operatorname{erfc}(x)=\frac{e^{-x^2}}{\sqrt{\pi}}\left(\frac{1}{2 x}-\frac{1}{4 x^3}-\int_x^{\infty} \frac{3 \exp \left(-t^2\right)}{2 t^4} d t\right), \]
Continuing with successive integration by parts we will obtain the asymptotic expansion.
5 Stirling’s formula
Use Laplace’s method to derive Stirling’s formula
\[ n ! \sim \sqrt{2 \pi n} n^n e^{-n} \quad n \rightarrow \infty \]
using the Gamma function. Also find the next term in the asymptotic series.
Gamma function, which is related to the factorial by \(\Gamma(n+1) = n!\) is defined as:
\[ \Gamma(n) = \int_0^{\infty} e^{-t} t^{n-1} dt \]
For large \(n\), we approximate this integral using Laplace’s method, which is effective for integrals of the form \(\int e^{M f(t)} dt\) where \(M\) is a large parameter. Here, we can rewrite the Gamma function as:
\[ \Gamma(n+1) = \int_0^{\infty} e^{n \log t - t} dt \]
Now, to apply Laplace’s method, we find the maximum of the function \(f(t) = \log t - \frac{t}{n}\). Taking the derivative and setting it to zero gives:
\[ \frac{d}{dt} \left( \log t - \frac{t}{n} \right) = \frac{1}{t} - \frac{1}{n} = 0 \]
This yields \(t = n\) as the point where \(f(t)\) attains its maximum. We then expand \(f(t)\) around this point:
\[ f(t) \approx f(n) + \frac{1}{2} f''(n) (t - n)^2 \]
where \(f(n) = \log n - 1\) and \(f''(n) = -1/n^2\). The integral becomes:
\[ \Gamma(n+1) \approx e^{n \log n - n} \int_{0}^{\infty} e^{-\frac{1}{2} \frac{(t-n)^2}{n}} dt \]
Changing variables with \(u = \frac{t-n}{\sqrt{n}}\) gives:
\[ \Gamma(n+1) \approx e^{n \log n - n} \sqrt{n} \int_{0}^{\infty} e^{-\frac{1}{2} u^2} du \]
Recognizing the Gaussian integral, we get:
\[ \Gamma(n+1) = n! \approx e^{n \log n - n} \sqrt{2 \pi n} = \sqrt{2 \pi n} n^n e^{-n} \quad n \rightarrow \infty \]
The second derivative \(f''(t)\) at the point \(t = n\) gives us the next leading term. The third derivative of \(f(t)\) is:
\[ f'''(t) = \frac{2}{t^3} \]
Evaluating this at \(t = n\) gives:
\[ f'''(n) = \frac{2}{n^3} \]
The corresponding term in the expansion is of the order \(\frac{1}{n}\), which leads to the next term in the series being \(\frac{1}{12n}\). Therefore, the improved Stirling’s formula, including the next term in the asymptotic series, is:
\[ n! \sim \sqrt{2 \pi n}\, n^n e^{-n} \left( 1 + \frac{1}{12n} \right) \]
6 Leading asymptotics of \(I(x)=\int_{-1}^1 e^{-x \sin ^4 t} d t\)
Use Laplace’s method to find the leading asymptotics of
\[ I(x)=\int_{-1}^1 e^{-x \sin ^4 t} d t \]
as \(x \rightarrow \infty\).
For the given integral, we observe that \(\sin^4 t\) is maximized when \(t = 0\) within the interval \([-1, 1]\). Near this point, the function \(\sin^4 t\) can be approximated by its Taylor expansion:
\[ \sin^4 t \approx t^4 \]
for small values of \(t\). The integral becomes:
\[ I(x) \approx \int_{-1}^1 e^{-x t^4} dt \]
As \(x \rightarrow \infty\), the contribution to the integral from regions where \(t\) is not very small becomes negligible, so we can extend the limits of the integral to infinity for the purpose of asymptotic approximation:
\[ I(x) \approx \int_{-\infty}^{\infty} e^{-x t^4} dt \]
Let \(u = x^{1/4} t\), then \(dt = x^{-1/4} du\) and the integral becomes:
\[ I(x) \approx \int_{-\infty}^{\infty} e^{-u^4} x^{-1/4} du \]
The integral \(\int_{-\infty}^{\infty} e^{-u^4} du\) is a constant (independent of \(x\)), so the leading asymptotic behavior of \(I(x)\) as \(x \rightarrow \infty\) is given by:
\[ I(x) \sim C x^{-1/4} \]
where \(C\) is the value of the integral \(\int_{-\infty}^{\infty} e^{-u^4} du\), which can be evaluated numerically.