Checking convergence of Gaussian integrals

[JD_PM]

Homework Statement

Given


$$Z(\lambda) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} dx \exp\left( -\frac{x^2}{2!}-\frac{\lambda}{4!}x^4\right)$$



a) What is the range of $\lambda$ such that $Z(\lambda)$ converges?



b) Find a compact expression for the series expansion of $Z(\lambda)$ for small $\lambda$ of the form



$$Z_N(\lambda) = \sum_{n=0}^N c_n \lambda^n$$



Is your series convergent? What is the radius of convergence?



c) Find a series expansion for $Z(\lambda)$ for $\lambda >> 1$ of the form



$$\hat Z_N(\lambda) = \sum_{n=0}^N d_n \lambda^{\left(2n+1 \right)/4}$$



Is this series convergent? For what value of $N$ will you obtain a value for $\hat Z_N(0,1)$ which is close to the exact value $Z(0,1)$.

Relevant Equations

N/A

a) First off, I computed the integral

\begin{align*}
Z(\lambda) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} dx \exp\left( -\frac{x^2}{2!}-\frac{\lambda}{4!}x^4\right) \\
&= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} dx \exp\left( -\frac{x^2}{2!}\right) \exp\left( -\frac{\lambda}{4!}x^4\right) \\
&= \frac{1}{\sqrt{2\pi}} \sqrt{2\pi} \left( \frac{(24)^{1/4}}{2(\lambda)^{1/4}} \Gamma\left( \frac 1 4 \right) \right) = \frac{(24)^{1/4}}{2(\lambda)^{1/4}} \Gamma\left( \frac 1 4 \right)
\end{align*}

So I would say that the range of convergence is $\lambda \in (0, +\infty]$, am I right?

For b) and c) I am quite confused. I have been trying to naively apply the series expansion for the exponential i.e.

$$Z(\lambda) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} dx \exp\left( -\frac{x^2}{2!}-\frac{\lambda}{4!}x^4\right) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} dx \sum_{n=0}^{\infty}\left( -\frac{x^2}{2!}-\frac{\lambda}{4!}x^4\right)^n / n! \tag{*}$$

But I do not see how (*) could lead to get b) and c) expressions; could you please give me a hint

Thank you

 

[mitochan]

a) You cannot integral $e^{-x^2}$ term and $e^{-x^4}$ term separately.

b) Try Taylor expansion of Z by $\lambda$. We can get $Z^{(n)}(0)$ easily.

 

[JD_PM]

a) You cannot integral $e^{-x^2}$ term and $e^{-x^4}$ term separately.

I see. How should I approach it then? I have been looking at Gaussian integral formulas; this is the best I could find

None of these fit the given integral though.

b) Try Taylor expansion of Z by $\lambda$. We can get $Z^{(n)}(0)$ easily.

\begin{align*}
Z(\lambda) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} dx \exp\left( -\frac{x^2}{2!}-\frac{\lambda}{4!}x^4\right) \\
&= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} dx \sum_{n=0}^{\infty}\left( -\frac{x^2}{2!}-\frac{\lambda}{4!}x^4\right)^n / n! \\
&\sim \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} dx \left[+1 -\frac{x^2}{2!}-\frac{\lambda}{4!}x^4 \right]
\end{align*}

So I would say that

\begin{equation*}
Z^{(n)}(0) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} dx \sum_{n=0}^{\infty}\left( -\frac{x^2}{2!}\right)^n / n!
\end{equation*}

But I do not see how the above will lead to get

$$Z^{(n)}(\lambda) = \sum_{n=0}^N c_n \lambda^n$$

 

[mitochan]

b) Try differentiating Z by $\lambda$. You will get
Z^{(1)}(0)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} e^{-x^2/2}x^4(-\frac{1}{24} )dx
Z^{(n)}(0)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} e^{-x^2/2}x^{4n}(-\frac{1}{24} )^n dx
You can get these values from the formula you found for a).

 

[Dr Transport]

Why don't you complete the square and try again.

 

[JD_PM]

b) Try differentiating Z by $\lambda$. You will get
Z^{(1)}(0)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} e^{-x^2/2}x^4(-\frac{1}{24} )dx
Z^{(n)}(0)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} e^{-x^2/2}x^{4n}(-\frac{1}{24} )^n dx
You can get these values from the formula you found for a).

Alright, I get your results. Then we have

$$Z^{(n)}(\lambda)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} e^{-x^2/2 -\lambda/4! x^4}x^{4n}\left(-\frac{1}{24} \right)^n dx \tag{**}$$

But how to get $Z^{(n)}(\lambda) = \sum_{n=0}^N c_n \lambda^n$ out of $(**)$?

Why don't you complete the square and try again.

Should I try

$$Z(\lambda) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} dx \exp\left[ -\left( x^2\sqrt{\frac{\lambda}{4!}} +\frac{x}{\sqrt{2}}\right)^2 + 2x^3 \sqrt{\frac{\lambda}{4!2}} \right]$$

With a change of variables $u := x^2\sqrt{\frac{\lambda}{4!}} +\frac{x}{\sqrt{2}}$ ?

 

[Dr Transport]

I was thinking complete the square in terms of x^2, not in terms of adding a x^3 term but a constant term more like let y^2 = (a + bx^2)^2

 

[JD_PM]

I was thinking complete the square in terms of x^2, not in terms of adding a x^3 term but a constant term more like let y^2 = (a + bx^2)^2

Then I am afraid I do see the form you have in mind... might you please give me the explicit form, so that I can think how to solve it?

 

[Dr Transport]

Then I am afraid I do see the form you have in mind... might you please give me the explicit form, so that I can think how to solve it?

your job, it's your homework. It took me about 5 lines to do by myself.

 

[JD_PM]

your job, it's your homework. It took me about 5 lines to do by myself.

You are right, let me try again. Did you solve the following integral

$$Z(\lambda) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} dx \exp\left[ -\left( x^2\sqrt{\frac{\lambda}{4!}} + \sqrt{\frac{3}{\lambda}}\right)^2 +\frac{3}{\lambda} \right]$$

With a change of variables $u := x^2\sqrt{\frac{\lambda}{4!}} + \sqrt{\frac{3}{\lambda}}$ ?

 

[Dr Transport]

Exactly... This should be tractable...

 

[Fred Wright]

Have you considered expanding $e^{-\frac{\lambda x^4}{4!}}$ and integrating term by term against $e^{\frac{-x^2}{2}}$?

 

[Dr Transport]

The original form of the integral is exact, I found it in Gradshteyn and Ryzhik this afternoon after I started to look at the problem more closely.

 

[mitochan]

But I do not see how the above will lead to get

You seem to confuse $Z_N$ Taylor series up to order N with $Z^{(n)}$ derivatives. $c_n$ is made from $Z^{(n)} (0)$.

 

[JD_PM]

Exactly... This should be tractable...

Alright so we have (I missed a 2 at #10)

$$Z(\lambda) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} dx \exp\left[ -\left( x^2\sqrt{\frac{\lambda}{4!}} + \sqrt{\frac{3}{2\lambda}}\right)^2 +\frac{3}{2\lambda} \right]$$

Labelling

$$a:= \sqrt{\frac{\lambda}{4!}}, \qquad b:= \sqrt{\frac{3}{2\lambda}}, \qquad c:= \frac{3}{2\lambda}$$

We end up with $1 / \sqrt{2\pi} \exp(c) \int_{-\infty}^{\infty} \exp(-(ax^2 +b)^2) dx$. I thus got the form you suggested. Then I would use the change of variables $y^2 = (ax^2 +b)^2 \iff y dy = 2ax(ax^2 +b)dx$.

But how to integrate

$$\frac{1}{\sqrt{2\pi}} e^c \int_{-\infty}^{\infty} e^{-y^2} \frac{1}{2a}\sqrt{\frac{a}{y-b}} dy$$

?

 

[Dr Transport]

If you expand the e^{-\frac{\lambda}{4!}x^4} term, you do not have to complete the square to integrate the results term by term.

 

[JD_PM]

Sorry @Dr Transport I am a bit confused; are you suggesting to abandon the 'complete the square' method and go for a different method (the one suggested by @Fred Wright ) ?

 

[JD_PM]

@Dr Transport , @Fred Wright thank you, I start to understand! I get (let me drop out the normalization factor for now)

\begin{align*}
\int_{-\infty}^{\infty} dx \exp\left( -\frac{x^2}{2!} - \frac{\lambda}{4!}x^4 \right) &= \int_{-\infty}^{\infty} dx \exp\left( -\frac{x^2}{2!} - \frac{\lambda}{4!}x^4 \right) = \int_{-\infty}^{\infty} dx \exp\left( -\frac{x^2}{2!} \right) \exp\left( - \frac{\lambda}{4!}x^4 \right) \\
&= \int_{-\infty}^{\infty} dx \exp\left( -\frac{x^2}{2!} \right) \left[ 1 - \frac{\lambda}{4!}x^4 + \frac{\lambda^2}{1152}x^8 - \ ... \ \left(- \frac{\lambda}{4!} x^4\right)^n \frac{1}{n!} \right] \\
&= \int_{-\infty}^{\infty} dx \exp\left( -\frac{x^2}{2!} \right) - \frac{\lambda}{4!}\int_{-\infty}^{\infty} dx \exp\left( -\frac{x^2}{2!} \right)x^4 + \frac{\lambda^2}{1152}\int_{-\infty}^{\infty} dx \exp\left( -\frac{x^2}{2!} \right)x^8 - \ ... \ + \left(-\frac{\lambda}{4!}\right)^n \int_{-\infty}^{\infty} dx \exp\left( -\frac{x^2}{2!} \right) \frac{x^{4n}}{n!}\\
&= \sqrt{2 \pi} - \frac{\lambda}{2^{5/2}} \sqrt{\pi} + \frac{35}{3 \times 2^{13/2}} \sqrt{\pi} \lambda^2 - \ ... \ \text{I do not see the pattern} \\
&= \sum_{n=0}^N c_n \lambda^n
\end{align*}

Where the coefficients $c_n$ are given, as @mitochan stated, by the derivatives $Z^{(n)}$

Next for me to understand is the range of convergence of this integral. Maybe I need to see the result of the the n-th integral first. I suspect that it should look similar to what follows

But instead of $x^{2n}$ we have $x^{4n}$.

Might you please shed some light on how to compute the n-th integral and then on the range of convergence of the general integral?

 

[vela]

Might you please shed some light on how to compute the n-th integral and then on the range of convergence of the general integral?

I'm not sure exactly what you are looking for. You already have all the pieces in your post. (Don't multiply all the constants out as you seem to be doing.)

Use something like the ratio test to find the radius of convergence.

 

[Dr Transport]

4n = 2(2n) , think about it...

 

[mitochan]

@JD_PM　You see the pattern
c_n=\frac{Z^{(n)}(0)}{n!}
=\frac{(-)^n}{\sqrt{2\pi}}\frac{2}{(4!)^n n!}\int_{0}^{+\infty}x^{4n} e^{-\frac{x^2}{2}}dx=\frac{(-)^n}{\sqrt{\pi}}\frac{2^{2n+1}}{(4!)^n n!}\int_{0}^{+\infty}x^{4n} e^{-x^2}dx＝...
c_0=1

 

[JD_PM]

Thank you all! I've been thinking and I would say I got section b) Let me argue it

\begin{align*}
Z(\lambda) &= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} dx e^{-x^2/2!} e^{-\lambda x^4/4!} \\
&= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} dx e^{-x^2/2!} \left( 1 - \frac{\lambda}{4!}x^4 + \ ... \ + \frac{(-)^n}{N!}\frac{\lambda^N}{(4!)^N}x^{4N} + ...\right) \\
&= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} dx e^{-x^2/2!} \sum_{n=0}^N \frac{(-)^n}{n!}\frac{\lambda^n}{(4!)^n}x^{4n} \\
&= \frac{1}{\sqrt{2 \pi}} \sum_{n=0}^N \frac{(-)^n}{n!}\frac{\lambda^n}{(4!)^n} \left( \int_{-\infty}^{\infty} x^{4n} e^{-x^2/2} dx\right)
\end{align*}

Solving the integral (where I used the change of variables $u:= x^2 / 2$ and looked up the following integral $\int_{0}^{\infty} e^{-n} x^{n-1} dx = \sqrt{n}$)

\begin{align*}
I= \int_{-\infty}^{\infty} x^{4n} e^{-x^2/2} dx &= 2\int_{0}^{\infty} x^{4n} e^{-x^2/2} dx \\
&= 2 \int_{0}^{\infty} (\sqrt{2t})^{4n} e^{-t} \frac{dt}{\sqrt{2t}} \\
&= 2 (\sqrt{2})^{4n - 1} \int_{0}^{\infty} t^{2n - 1/2} e^{-t} dt \\
&= (\sqrt{2})^{4n + 1} \sqrt{2n + \frac 1 2}
\end{align*}

We end up with

\begin{align*}
Z(\lambda) &= \frac{1}{\sqrt{2 \pi}} \sum_{n=0}^N \frac{(-)^n}{n!}\frac{1}{(4!)^n} (\sqrt{2})^{4n + 1} \sqrt{2n + \frac 1 2} \lambda^n \\
&= \sum_{n=0}^N c_n \lambda^n \\
\end{align*}

Mmm but I get $c_0 = 1 / \sqrt{2 \pi}$ instead of $c_0 = 1$. So I guess I missed a $\sqrt{2 \pi}$ term when computing $I$ but I still do not see it though.

Use something like the ratio test to find the radius of convergence.

Thanks, using the ratio test I get that the integral is convergent and its radius of convergence $R = \infty$. However, given that my computation of $I$ might be wrong, it may be wrong.

=\frac{(-)^n}{\sqrt{2\pi}}\frac{2}{(4!)^n n!}\int_{0}^{+\infty}x^{4n} e^{-\frac{x^2}{2}}dx=\frac{(-)^n}{\sqrt{\pi}}\frac{2^{2n+1}}{(4!)^n n!}\int_{0}^{+\infty}x^{4n} e^{-x^2}dx＝...

Thanks but might you shed some light on how did you go from the LHS to the RHS?

 

[mitochan]

You should take better care of integration
\int_0^{+\infty} t^{2n-\frac{1}{2}}e^{-t}dt.
By partial integration
\int_0^{+\infty} t^{2n-\frac{1}{2}}e^{-t}dt=[ -t^{2n-\frac{1}{2}}e^{-t}]_0^{+\infty} - (-)(2n-\frac{1}{2})\int_0^{+\infty}t^{2n-\frac{1}{2}-1}e^{-t}dt=(2n-\frac{1}{2})\int_0^{+\infty}t^{2n-\frac{1}{2}-1}e^{-t}dt
= (2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})...\frac{3}{2}\frac{1}{2} \int_0^{+\infty}t^{-\frac{1}{2}}e^{-t}dt=(2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})...\frac{3}{2}\frac{1}{2}\ \ 2\int_0^{+\infty}e^{-t}d(\sqrt{t})

 

[Dr Transport]

\int_0^\infty x^{2n}e^{-ax^2}dx = \frac{1*3*5...(2n-1)}{2^{n+1}a^n}\sqrt{\frac{\pi}{a}}

figure it out from here n \to 2n

 

[JD_PM]

My apologies for the late reply

You should take better care of integration
\int_0^{+\infty} t^{2n-\frac{1}{2}}e^{-t}dt.
By partial integration
\int_0^{+\infty} t^{2n-\frac{1}{2}}e^{-t}dt=[ -t^{2n-\frac{1}{2}}e^{-t}]_0^{+\infty} - (-)(2n-\frac{1}{2})\int_0^{+\infty}t^{2n-\frac{1}{2}-1}e^{-t}dt=(2n-\frac{1}{2})\int_0^{+\infty}t^{2n-\frac{1}{2}-1}e^{-t}dt
= (2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})...\frac{3}{2}\frac{1}{2} \int_0^{+\infty}t^{-\frac{1}{2}}e^{-t}dt=(2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})...\frac{3}{2}\frac{1}{2}\ \ 2\int_0^{+\infty}e^{-t}d(\sqrt{t})

Indeed! I have noticed I made a mistake at #22. It is of course not a square root but a gamma function!\begin{align*}
I= \int_{-\infty}^{\infty} x^{4n} e^{-x^2/2} dx &= 2\int_{0}^{\infty} x^{4n} e^{-x^2/2} dx \\
&= 2 \int_{0}^{\infty} (\sqrt{2t})^{4n} e^{-t} \frac{dt}{\sqrt{2t}} \\
&= 2 (\sqrt{2})^{4n - 1} \int_{0}^{\infty} t^{2n - 1/2} e^{-t} dt \\
&= (\sqrt{2})^{4n - 1} \Gamma \left( 2n + \frac 1 2 \right) \\
&= 2 (\sqrt{2})^{4n - 1} \frac{4n! \sqrt{\pi}}{2^{4n}(2n)!}
\end{align*}

This is how I got the last equality.

Hence we conclude that

\begin{align*}
Z(\lambda) &= \frac{1}{\sqrt{2 \pi}} \sum_{n=0}^N \frac{(-)^n}{n!}\frac{1}{(4!)^n} (\sqrt{2})^{4n - 1} \Gamma\left(2n + \frac 1 2 \right) \lambda^n \\
&= \sum_{n=0}^N c_n \lambda^n \\
\end{align*}

 

[JD_PM]

Let us tackle c) now.

c) Find a series expansion for $Z(\lambda)$ for $\lambda >> 1$ of the form
$$\hat Z_N(\lambda) = \sum_{n=0}^N d_n \lambda^{\left(2n+1 \right)/4}$$
Is this series convergent? For what value of $N$ will you obtain a value for $\hat Z_N(0,1)$ which is close to the exact value $Z(0,1)$.

I am a bit confused here. How should we massage the original $Z(\lambda)$ so that we get the form $\hat Z_N(\lambda)$ has?

 

[mitochan]

You can write down $c_n$ simply e.g. with no $\sqrt{\pi}$.

 

[JD_PM]

You can write down $c_n$ simply e.g. with no $\sqrt{\pi}$.

Indeed, by means of

$$\displaystyle \Gamma\left(2n+ \frac 1 2 \right) = \frac{(4n)!\sqrt{\pi}}{2^{4n}(2n)!}$$

Any tips on how to approach c)? :) I am currently trying to find a method.

 

[mitochan]

Why don't you continue calculation of post #23. I did not complete it because of homework policy.

 

[JD_PM]

You should take better care of integration
\int_0^{+\infty} t^{2n-\frac{1}{2}}e^{-t}dt.
By partial integration
\int_0^{+\infty} t^{2n-\frac{1}{2}}e^{-t}dt=[ -t^{2n-\frac{1}{2}}e^{-t}]_0^{+\infty} - (-)(2n-\frac{1}{2})\int_0^{+\infty}t^{2n-\frac{1}{2}-1}e^{-t}dt=(2n-\frac{1}{2})\int_0^{+\infty}t^{2n-\frac{1}{2}-1}e^{-t}dt
= (2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})...\frac{3}{2}\frac{1}{2} \int_0^{+\infty}t^{-\frac{1}{2}}e^{-t}dt=(2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})...\frac{3}{2}\frac{1}{2}\ \ 2\underbrace{\int_0^{+\infty}e^{-t}d(\sqrt{t})}_{I_1}

I agree, let's finish the above computation first ;)

We see that $(2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})\cdots\frac{3}{2}\frac{1}{2} = \Gamma\left(2n+ \frac 1 2 \right)$ . Now we should introduce a change of variables (I am not used to work with differentials with a square root, so I might have done a mistake). Setting $x^2:=t$ we get

$$\displaystyle I_1 = 2 \int_{0}^{\infty} x^2 e^{-x^2}\,{dx}$$

Now, there are several methods to solve this integral. I particularly liked the one involving differentiation of $e^{-x^2}$

$$\frac{d^2}{dx^2}e^{-x^2}=-2e^{-x^2}+4x^2e^{-x^2} \tag{*}$$Taking the integral of $(*)$'s LHS yields

$$\int_{0}^{\infty}\left(\frac{d^2}{dx^2}e^{-x^2}\right)dx=\frac{d^2}{dx^2}\int_{0}^{\infty} e^{-x^2}dx=\frac{d^2}{dx^2}\frac{\sqrt\pi}{2}=0$$

Taking the integral of $(*)$'s RHS yields\begin{align*}
\int_{0}^{\infty} \left( -2e^{-x^2}+4x^2e^{-x^2}\right) dx = 0 \Rightarrow \int_{0}^{\infty} x^2 e^{-x^2}\,{dx} = \sqrt{\pi}/4
\end{align*}

Hence $\displaystyle I_1 = \sqrt{\pi}/2$ So

\begin{align*}
&(2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})...\frac{3}{2}\frac{1}{2}\ \ 2\underbrace{\int_0^{+\infty}e^{-t}d(\sqrt{t})}_{I_1} \\
&= \sqrt{\pi} \cdot (2n-\frac{1}{2}) \cdot (2n-\frac{3}{2}) \cdot (2n-\frac{5}{2})\cdots\frac{3}{2} \cdot \frac{1}{2} \\
&= \frac{(4n)!\sqrt{\pi}}{2^{4n}(2n)!}
\end{align*}

Which leads to the expected final answer, shown #25

 

[mitochan]

So you are prepared to write down $c_n=...$.

 

[JD_PM]

So you are prepared to write down $c_n=...$.

Indeed (I would say it cannot be simplified any further)

\begin{equation*}
c_n = \frac{1}{\sqrt{2}} \sum_{n=0}^N\frac{(-)^n}{n!}\frac{1}{(4!)^n} (\sqrt{2})^{4n - 1} \frac{(4n)!}{2^{4n}(2n)!}
\end{equation*}

 

[mitochan]

No summation and take a look at factor $2^n$.

 

[JD_PM]

No summation and take a look at factor $2^n$.

Oops my bad. It can be simplified further

\begin{equation*}
c_n = \frac{1}{2} \sum_{n=0}^N\frac{(-)^n}{n!}\frac{1}{(4!)^n} \frac{(4n)!}{2^{3n}(2n)!}
\end{equation*}

 

[mitochan]

Do it more carefully.

 

[JD_PM]

Do it more carefully.

\begin{align*}
c_n &= \frac{1}{\sqrt{2}}\frac{(-)^n}{n!}\frac{1}{(4!)^n}(2)^{2n}\frac{1}{\sqrt{2}} \frac{(4n)!}{2^{4n}(2n)!} \\
&= \frac{1}{2} \frac{(-)^n}{n!}\frac{1}{(4!)^n} \frac{(4n)!}{2^{2n}(2n)!}
\end{align*}

 

[mitochan]

Thanks. I got c0=1 in post #21 and your result shows c0=1/2. I should appreciate it if you would check factor 2 missing or duplicating.

 

[JD_PM]

Let me summarize what we are doing. At #21 you obtained (let me plug the solution of the integral)

\begin{align*}
c_n &= \frac{(-)^n}{\sqrt{2\pi}}\frac{2}{(4!)^n n!}\int_{0}^{+\infty}x^{4n} e^{-\frac{x^2}{2}}dx=\frac{(-)^n}{\sqrt{\pi}}\frac{2^{2n+1}}{(4!)^n n!}\int_{0}^{+\infty}x^{4n} e^{-x^2}dx \\
&= \frac{(-)^n 2^{2n+1}}{(4!)^n n!} (\sqrt{2})^{4n - 1} \frac{(4n)!}{2^{4n}(2n)!}.
\end{align*}

And I got

\begin{align*}
c_n &= \frac{1}{\sqrt{2}}\frac{(-)^n}{n!}\frac{1}{(4!)^n}(2)^{2n}\frac{1}{\sqrt{2}} \frac{(4n)!}{2^{4n}(2n)!} \\
&= \frac{1}{2} \frac{(-)^n}{n!}\frac{1}{(4!)^n} \frac{(4n)!}{2^{2n}(2n)!}
\end{align*}

It indeed seems I am missing a $2$ factor. I will recheck my computation.

 

[JD_PM]

OK Got it, thanks for your patience!

Indeed (I would say it cannot be simplified any further)

\begin{equation*}
c_n = \frac{1}{\sqrt{2}} \sum_{n=0}^N\frac{(-)^n}{n!}\frac{1}{(4!)^n} (\sqrt{2})^{4n - 1} \frac{(4n)!}{2^{4n}(2n)!}
\end{equation*}

I dropped a two (i.e. we have $\int_{-\infty}^{\infty} x^{4n} e^{-x^2/2} \, dx = 2 (\sqrt{2})^{4n - 1} \Gamma\left(2n+ \frac 1 2 \right)$ and not $\int_{-\infty}^{\infty} x^{4n} e^{-x^2/2} \, dx = (\sqrt{2})^{4n - 1} \Gamma\left(2n+ \frac 1 2 \right)$).

So now my answer matches yours i.e.

\begin{align*}
c_n &= \frac{2}{\sqrt{2}}\frac{(-)^n}{n!}\frac{1}{(4!)^n}(2)^{2n}\frac{1}{\sqrt{2}} \frac{(4n)!}{2^{4n}(2n)!} \\
&= \frac{(-)^n}{n!}\frac{1}{(4!)^n} \frac{(4n)!}{2^{2n}(2n)!}
\end{align*}

 

[mitochan]

Thanks. So we can proceed to investigate the series converges or diverges.

 

[JD_PM]

Thanks. So we can proceed to investigate the series converges or diverges.

I did this section some days ago via ratio test and got that the series was convergent, with radius of convergence $R=\infty$

Do you mind if we advance or do you want me to show it explicitly?

 

[mitochan]

If you do not mind please show your result.

 

[JD_PM]

If you do not mind please show your result.

Using the ratio test we get

\begin{align*}
\lim_{n \to \infty} \left|\left( c_{n + 1} \frac{1}{c_n}\right)\right| &= \frac{(-)^{n+1}}{(n+1)!} \frac{1}{(4!)^{n+1}} \frac{\left[4(n+1) \right]!}{2^{2(n+1)}\left[ 2(n+1)\right]!} \\
&\times \frac{n!}{(-)^n}(4!)^n \frac{2^{2n}(2n)!}{(4n)!} \\
&= \lim_{n \to \infty} \left(\frac{1}{4! (n+1)}\right) \\
&= 0 < 1
\end{align*}

Hence $R= + \infty$

 

[mitochan]

I calculated it
\frac{c_{n+1}}{c_n}=\frac{-1}{4*4!}\frac{(4n+4)(4n+3)(4n+2)(4n+1)}{(n+1)(2n+2)(2n+1)}=\frac{-1}{4!}\frac{(4n+3)(4n+1)}{(n+1)}=-\frac{2}{3}\frac{(1+3/4n)(1+1/4n)}{1+1/n}n
Please check our results.

 

[JD_PM]

I calculated it
\frac{c_{n+1}}{c_n}=\frac{-1}{4*4!}\frac{(4n+4)(4n+3)(4n+2)(4n+1)}{(n+1)(2n+2)(2n+1)}=\frac{-1}{4!}\frac{(4n+3)(4n+1)}{(n+1)}=-\frac{2}{3}\frac{(1+3/4n)(1+1/4n)}{1+1/n}n
Please check our results.

You are right, my bad! (I have to acknowledge that I rushed, my apologies).

Let me be more careful

\begin{align*}
c_{n + 1} \frac{1}{c_n} &= \frac{(4n+4)!}{(n+1)!(4!)^{n+1} 2^{2n+2}(2n+2)!}\times \\
&\times \frac{n! (4!)^n 2^{2n}(2n)!}{(4n)!} \\
&= \frac{(4n+3)(4n+1)}{24(n+1)} \\
&= \frac{n}{24}\left( \frac{(4+3/n)(4+1/n)}{(4+1/n)} \right)
\end{align*}

Where, to get to the first equality, I used factorial properties and simplified

$$(4n+4)! = (4n+4)(4n+3)(4n+2)(4n+1)(4n)!$$

$$(n+1)! = (n+1)(n)!, \qquad (2n+2)! = (2n+2)(2n+1)(2n)!$$

So the series diverges!

\begin{equation*}
\lim_{n\to \infty} \left( \frac{n}{24}\left( \frac{(4+3/n)(4+1/n)}{(4+1/n)} \right) \right) = \infty
\end{equation*}

And the radius of convergence is given by

$$R=0$$

 

[JD_PM]

Regarding c). Let me check if I am on the right track

I would introduce a change of variables $u := \lambda^{1/4} x$ and expand one of the exponentials only so that $Z(\lambda)$ becomes

\begin{align*}
Z(\lambda) &= \frac{1}{\sqrt{2\pi}} \lambda^{-1/4} \int_{-\infty}^{\infty} du \exp\left( -\frac{u^2}{2!\sqrt{\lambda}}-\frac{1}{4!}u^4\right) \\
&= \frac{1}{\sqrt{2\pi}} \lambda^{-1/4} \sum_{n=0}^{\infty} \int_{-\infty}^{\infty} du e^{-u^4/4!} \left( \frac{-u^2}{2! \sqrt{\lambda}} \right)^n \frac{1}{n!}
\end{align*}Mmm so all would boil down to computing an integral of the form

\begin{align*}
\int_{-\infty}^{\infty} du \ e^{-u^4/4!} u^{2n}
\end{align*}

 

[mitochan]

Let me confirm you say

d_n=\frac{1}{\sqrt{2\pi}}\frac{(-)^n}{n!}\int_{-\infty}^{+\infty} e^{-\frac{u^4}{4!}}\frac{u^{2n}}{2^n} du

Why don't you try changing integral paremeters
\frac{u^4}{4!}=t
and do a similar job with b) ?

 

[JD_PM]

@mitochan I see how to complete the exercise, thank you for your help and patience