Challenge Math Challenge - May 2021

[kshitij]

You should have used the approximations that I gave in the problem statement, not a calculator so that the final conclusion would be

but, yes, this is correct.

The reason is: If you use a calculator, then you make implicitly the assumption, that it is more precise than the values you have been given. Well, this is probably correct, as long as you didn't use a slide rule. Nevertheless, it is an assumption about a device you have no control of and you should be aware of it, e.g. if you write a protocol of an experiment.

I'll keep that in mind next time

 

[fresh_42]

I'll keep that in mind next time

It is not important in a case like this. I still have the dream that people can learn from those problems now and then, and the problem was all about precision. Hence the lesson to be learned is, that the precision of a result is always as good as the precision of the measurement, or calculating devices is. My remark was meant to sharpen your senses, not to criticize anything.

 

[graphking]

7. Let $\alpha$ be an algebraic number of degree $n\geq 1.$ Then there is a real number $c>0$ such that for all $\mathbb{Q}\ni\dfrac{p}{q}\neq \alpha$
$$
\left|\alpha -\dfrac{p}{q}\right|\geq \dfrac{c}{q^n}
$$

Hi guys, I didn't read all your post but it seems that no one say anything about Q7. I remember that is a beautiful theorem called Liouville Theorem to describe algebraic number and transcendental number a little bit. Here's some hint:
let f(x) be the polynomial for $\alpha$ . consider the nearest p/q to $\alpha$ , when p is variable and q is settled. Then try f($\alpha$)-f(p/q), factorize it and get $\alpha - p/q$.

 

[julian]

I may have done it, must go to bed now:

The $\alpha$ is the root of an $n-$th order polynomial with integer coefficients:

\begin{align*}
f (x) = \sum_{i=0}^n a_i x^i
\end{align*}

Put

\begin{align*}
\tilde{\alpha} = \frac{p}{q} .
\end{align*}

If $f (x)$ has rational roots then denote them as $\{ r_1 , \dots , r_k \}$.

Case (a) $\tilde{\alpha} \not\in \{ r_1 , \dots , r_k \}$ and $|\alpha - \tilde{\alpha}| \leq 1$.

We have the identity

\begin{align*}
f (x) & = \sum_{i=1}^n a_i x^i - \sum_{i=1}^n a_i a^i
\nonumber \\
& = \sum_{i=1}^n a_i (x - a) (x^{i-1} + x^{i-2} a + \cdots + a^{i-1})
\nonumber \\
& = (x - a) \sum_{i=1}^n a_i \sum_{j=0}^{i-1} x^{i-1-j} a^j
\end{align*}

Application of the triangle inequality $|\tilde{\alpha}| = |\tilde{\alpha} - \alpha + \alpha| \leq |\alpha - \tilde{\alpha}| + |\alpha|$ together with $|\alpha - \tilde{\alpha}| \leq 1$ implies $|\tilde{\alpha}| \leq 1 + |\alpha|$. We use this to obtain the inequality:

\begin{align*}
|f (\alpha) - f (\tilde{\alpha})| & \leq |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i| \sum_{j=0}^{i-1} |\alpha^{i-1-j} \tilde{\alpha}^j |
\nonumber \\
& = |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i| \sum_{j=0}^{i-1} |\alpha^{i-1}| | \alpha^{-j} (1 + |\alpha|)^j |
\nonumber \\
& = |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i \alpha^{i-1}| \sum_{j=0}^{i-1} \left( 1 + \frac{1}{|\alpha|} \right)^j
\nonumber \\
& = |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i \alpha^{i-1}| \frac{\left( 1 + \frac{1}{|\alpha|} \right)^i - 1}{\left( 1 + \frac{1}{|\alpha|} \right) - 1}
\nonumber \\
& = |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i| \left( (|\alpha| + 1)^i - |\alpha|^i \right)
\nonumber \\
& = |\alpha - \tilde{\alpha}| C_\alpha
\end{align*}

where $C_\alpha = \sum_{i=1}^n |a_i| \left( (|\alpha| + 1)^i - |\alpha|^i \right) > 0$. We have

\begin{align*}
|f ( \tilde{\alpha} )| & = \left|\frac{a_0 q^n + a_1 q^{n-1} p + \cdots + a_n p^n}{q^n} \right| \geq \frac{1}{q^n}
\end{align*}

as the numerator is a nonzero integer, and so

\begin{align*}
|\alpha - \tilde{\alpha}| & \geq \frac{1}{C_\alpha} |f ( \tilde{\alpha} )| \geq \frac{1}{C_\alpha} \frac{1}{q^n}
\end{align*}

Case (b) $\tilde{\alpha} \not\in \{ r_1 , \dots , r_k \}$ and $|\alpha - \tilde{\alpha}| > 1$.

\begin{align*}
|\alpha - \tilde{\alpha}| & > 1 \geq \frac{1}{q^n}
\end{align*}

Case (c) $\alpha \not\in \{ r_1 , \dots , r_k \}$, $\tilde{\alpha} \in \{ r_1 , \dots , r_k \}$.

Choose $C_r = \min_i | \alpha - r_i| > 0$. Then

\begin{align*}
|\alpha - \tilde{\alpha}| & \geq C_r \geq \frac{C_r}{q^n}
\end{align*}

(If there are no rational roots, ignore case (c)).

Case (d) Say $\alpha \in \{ r_1 , \dots , r_k \}$ and $\tilde{\alpha} \in \{ r_1 , \dots , r_k \}$ (in which case $k > 1$). Set $\overline{C}_r = \min_{i \not= j}|r_i - r_j| > 0$,

\begin{align*}
|\alpha - \tilde{\alpha}| & \geq \overline{C}_r \geq \overline{C}_r \frac{1}{q^n}
\end{align*}

(If there are no rational roots, ignore case (d)).

Finally, we can write

\begin{align*}
c = \min \left\{
\begin{matrix}
\frac{1}{C_\alpha} & : |\alpha - \frac{p}{q}| \leq 1, \quad \frac{p}{q} \not\in \{ r_1 , \dots , r_k \} \\
1 & : |\alpha - \frac{p}{q}| > 1, \quad \frac{p}{q} \not\in \{ r_1 , \dots , r_k \} \\
C_r & : \alpha \not\in \{ r_1 , \dots , r_k \} , \quad \frac{p}{q} \in \{ r_1 , \dots , r_k \} \\
\overline{C}_r & : \alpha , \frac{p}{q} \in \{ r_1 , \dots , r_k \}
\end{matrix}
\right.
\end{align*}

then

\begin{align*}
\left| \alpha - \frac{p}{q} \right| & \geq \frac{c}{q^n}
\end{align*}

where $c > 0$.

 

[fresh_42]

I may have done it, must go to bed now:

The $\alpha$ is the root of an $n-$th order polynomial with integer coefficients:

\begin{align*}
f (x) = \sum_{i=0}^n a_i x^i
\end{align*}

Put

\begin{align*}
\tilde{\alpha} = \frac{p}{q} .
\end{align*}

If $f (x)$ has rational roots then denote them as $\{ r_1 , \dots , r_k \}$.

Case (a) $\tilde{\alpha} \not\in \{ r_1 , \dots , r_k \}$ and $|\alpha - \tilde{\alpha}| \leq 1$.

We have the identity

\begin{align*}
f (x) & = \sum_{i=1}^n a_i x^i - \sum_{i=1}^n a_i a^i
\nonumber \\
& = \sum_{i=1}^n a_i (x - a) (x^{i-1} + x^{i-2} a + \cdots + a^{i-1})
\nonumber \\
& = (x - a) \sum_{i=1}^n a_i \sum_{j=0}^{i-1} x^{i-1-j} a^j
\end{align*}

The $a$ on the right-hand side should be an $\alpha .$

Application of the triangle inequality $|\tilde{\alpha}| = |\tilde{\alpha} - \alpha + \alpha| \leq |\alpha - \tilde{\alpha}| + |\alpha|$ together with $|\alpha - \tilde{\alpha}| \leq 1$ implies $|\tilde{\alpha}| \leq 1 + |\alpha|$. We use this to obtain the inequality:

\begin{align*}
|f (\alpha) - f (\tilde{\alpha})| & \leq |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i| \sum_{j=0}^{i-1} |\alpha^{i-1-j} \tilde{\alpha}^j |
\nonumber \\
& = |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i| \sum_{j=0}^{i-1} |\alpha^{i-1}| | \alpha^{-j} (1 + |\alpha|)^j |
\nonumber \\
& = |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i \alpha^{i-1}| \sum_{j=0}^{i-1} \left( 1 + \frac{1}{|\alpha|} \right)^j
\nonumber \\
& = |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i \alpha^{i-1}| \frac{\left( 1 + \frac{1}{|\alpha|} \right)^i - 1}{\left( 1 + \frac{1}{|\alpha|} \right) - 1}
\nonumber \\
& = |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i| \left( (|\alpha| + 1)^i - |\alpha|^i \right)
\nonumber \\
& = |\alpha - \tilde{\alpha}| C_\alpha
\end{align*}

The first equality sign has to be less or equal.

where $C_\alpha = \sum_{i=1}^n |a_i| \left( (|\alpha| + 1)^i - |\alpha|^i \right) > 0$. We have

\begin{align*}
|f ( \tilde{\alpha} )| & = \left|\frac{a_0 q^n + a_1 q^{n-1} p + \cdots + a_n p^n}{q^n} \right| \geq \frac{1}{q^n}
\end{align*}

as the numerator is a nonzero integer, and so

\begin{align*}
|\alpha - \tilde{\alpha}| & \geq \frac{1}{C_\alpha} |f ( \tilde{\alpha} )| \geq \frac{1}{C_\alpha} \frac{1}{q^n}
\end{align*}

Case (b) $\tilde{\alpha} \not\in \{ r_1 , \dots , r_k \}$ and $|\alpha - \tilde{\alpha}| > 1$.

\begin{align*}
|\alpha - \tilde{\alpha}| & > 1 \geq \frac{1}{q^n}
\end{align*}

Sloppy, since $q^n < 0$ isn't ruled out and $c$ cannot swallow the sign.

Case (c) $\alpha \not\in \{ r_1 , \dots , r_k \}$, $\tilde{\alpha} \in \{ r_1 , \dots , r_k \}$.

I assume this is meant to be the other way around since we already covered all cases of $\alpha \not\in \{ r_1 , \dots , r_k \}.$

The good news is that it is irrelevant because we may assume $f(x)$ to be irreducible over $\mathbb{Q}$.

Choose $C_r = \min_i | \alpha - r_i| > 0$. Then

\begin{align*}
|\alpha - \tilde{\alpha}| & \geq C_r \geq \frac{C_r}{q^n}
\end{align*}

(If there are no rational roots, ignore case (c)).

Case (d) Say $\alpha \in \{ r_1 , \dots , r_k \}$ and $\tilde{\alpha} \in \{ r_1 , \dots , r_k \}$ (in which case $k > 1$). Set $\overline{C}_r = \min_{i \not= j}|r_i - r_j| > 0$,

\begin{align*}
|\alpha - \tilde{\alpha}| & \geq \overline{C}_r \geq \overline{C}_r \frac{1}{q^n}
\end{align*}

(If there are no rational roots, ignore case (d)).

Finally, we can write

\begin{align*}
c = \min \left\{
\begin{matrix}
\frac{1}{C_\alpha} & : |\alpha - \frac{p}{q}| \leq 1, \quad \frac{p}{q} \not\in \{ r_1 , \dots , r_k \} \\
1 & : |\alpha - \frac{p}{q}| > 1, \quad \frac{p}{q} \not\in \{ r_1 , \dots , r_k \} \\
C_r & : \alpha \not\in \{ r_1 , \dots , r_k \} , \quad \frac{p}{q} \in \{ r_1 , \dots , r_k \} \\
\overline{C}_r & : \alpha , \frac{p}{q} \in \{ r_1 , \dots , r_k \}
\end{matrix}
\right.
\end{align*}

then

\begin{align*}
\left| \alpha - \frac{p}{q} \right| & \geq \frac{c}{q^n}
\end{align*}

where $c > 0$.

It would have been shorter to write $f(x)=(x-\alpha )g(x)\in \mathbb{C}[x]$ and then discuss the neighborhood of $\alpha$ by continuity of $g(x)$.

 

[julian]

The $a$ on the right-hand side should be an $\alpha .$

I meant to write $f(x) - f(a)$ on the LHS. I was just simply stating an identity, I make appropriate replacements for $x$ and $a$ in the next part.

The first equality sign has to be less or equal.

Typo.

I assume this is meant to be the other way around since we already covered all cases of $\alpha \not\in \{ r_1 , \dots , r_k \}.$

The good news is that it is irrelevant because we may assume $f(x)$ to be irreducible over $\mathbb{Q}$.

I thought I had to consider $\tilde{\alpha} \in \{ r_1 , \dots , r_k \}$ separately as the proof used in case (a) used that $f(\tilde{\alpha}) \not= 0$. But of course I didn't have to do this because, as you alluded to, if $p/q$ were a root then you could factor out $x - p/q$ from $f (x)$ and $\alpha$ would satisfy a polynomial with rational coefficients whose degree is less than $n$.

 

[fresh_42]

@julian, your proof was fine. My remarks' meaning was mainly to assure you, that I had actually read your proof, rather than criticism.

 

[graphking]

I may have done it, must go to bed now:

The $\alpha$ is the root of an $n-$th order polynomial with integer coefficients:

\begin{align*}
f (x) = \sum_{i=0}^n a_i x^i
\end{align*}

Put

\begin{align*}
\tilde{\alpha} = \frac{p}{q} .
\end{align*}

If $f (x)$ has rational roots then denote them as $\{ r_1 , \dots , r_k \}$.

Case (a) $\tilde{\alpha} \not\in \{ r_1 , \dots , r_k \}$ and $|\alpha - \tilde{\alpha}| \leq 1$.

We have the identity

\begin{align*}
f (x) & = \sum_{i=1}^n a_i x^i - \sum_{i=1}^n a_i a^i
\nonumber \\
& = \sum_{i=1}^n a_i (x - a) (x^{i-1} + x^{i-2} a + \cdots + a^{i-1})
\nonumber \\
& = (x - a) \sum_{i=1}^n a_i \sum_{j=0}^{i-1} x^{i-1-j} a^j
\end{align*}

Application of the triangle inequality $|\tilde{\alpha}| = |\tilde{\alpha} - \alpha + \alpha| \leq |\alpha - \tilde{\alpha}| + |\alpha|$ together with $|\alpha - \tilde{\alpha}| \leq 1$ implies $|\tilde{\alpha}| \leq 1 + |\alpha|$. We use this to obtain the inequality:

\begin{align*}
|f (\alpha) - f (\tilde{\alpha})| & \leq |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i| \sum_{j=0}^{i-1} |\alpha^{i-1-j} \tilde{\alpha}^j |
\nonumber \\
& = |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i| \sum_{j=0}^{i-1} |\alpha^{i-1}| | \alpha^{-j} (1 + |\alpha|)^j |
\nonumber \\
& = |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i \alpha^{i-1}| \sum_{j=0}^{i-1} \left( 1 + \frac{1}{|\alpha|} \right)^j
\nonumber \\
& = |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i \alpha^{i-1}| \frac{\left( 1 + \frac{1}{|\alpha|} \right)^i - 1}{\left( 1 + \frac{1}{|\alpha|} \right) - 1}
\nonumber \\
& = |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i| \left( (|\alpha| + 1)^i - |\alpha|^i \right)
\nonumber \\
& = |\alpha - \tilde{\alpha}| C_\alpha
\end{align*}

where $C_\alpha = \sum_{i=1}^n |a_i| \left( (|\alpha| + 1)^i - |\alpha|^i \right) > 0$. We have

\begin{align*}
|f ( \tilde{\alpha} )| & = \left|\frac{a_0 q^n + a_1 q^{n-1} p + \cdots + a_n p^n}{q^n} \right| \geq \frac{1}{q^n}
\end{align*}

as the numerator is a nonzero integer, and so

\begin{align*}
|\alpha - \tilde{\alpha}| & \geq \frac{1}{C_\alpha} |f ( \tilde{\alpha} )| \geq \frac{1}{C_\alpha} \frac{1}{q^n}
\end{align*}

Case (b) $\tilde{\alpha} \not\in \{ r_1 , \dots , r_k \}$ and $|\alpha - \tilde{\alpha}| > 1$.

\begin{align*}
|\alpha - \tilde{\alpha}| & > 1 \geq \frac{1}{q^n}
\end{align*}

Case (c) $\alpha \not\in \{ r_1 , \dots , r_k \}$, $\tilde{\alpha} \in \{ r_1 , \dots , r_k \}$.

Choose $C_r = \min_i | \alpha - r_i| > 0$. Then

\begin{align*}
|\alpha - \tilde{\alpha}| & \geq C_r \geq \frac{C_r}{q^n}
\end{align*}

(If there are no rational roots, ignore case (c)).

Case (d) Say $\alpha \in \{ r_1 , \dots , r_k \}$ and $\tilde{\alpha} \in \{ r_1 , \dots , r_k \}$ (in which case $k > 1$). Set $\overline{C}_r = \min_{i \not= j}|r_i - r_j| > 0$,

\begin{align*}
|\alpha - \tilde{\alpha}| & \geq \overline{C}_r \geq \overline{C}_r \frac{1}{q^n}
\end{align*}

(If there are no rational roots, ignore case (d)).

Finally, we can write

\begin{align*}
c = \min \left\{
\begin{matrix}
\frac{1}{C_\alpha} & : |\alpha - \frac{p}{q}| \leq 1, \quad \frac{p}{q} \not\in \{ r_1 , \dots , r_k \} \\
1 & : |\alpha - \frac{p}{q}| > 1, \quad \frac{p}{q} \not\in \{ r_1 , \dots , r_k \} \\
C_r & : \alpha \not\in \{ r_1 , \dots , r_k \} , \quad \frac{p}{q} \in \{ r_1 , \dots , r_k \} \\
\overline{C}_r & : \alpha , \frac{p}{q} \in \{ r_1 , \dots , r_k \}
\end{matrix}
\right.
\end{align*}

then

\begin{align*}
\left| \alpha - \frac{p}{q} \right| & \geq \frac{c}{q^n}
\end{align*}

where $c > 0$.

Yeah, that was what I mean! But as @fresh_42 has figured out, the f(x) is actually the minimal polynomial (minimal in the degree of polynomial) , that would make the proof shorter (like case c,d can be erased). Also, like I said:

consider the nearest p/q to α , when p is variable and q is settled.

so case b could be avoided. I think this quote is a first understanding of this problem, we should discover p is variable and q is settled.

by the way, @fresh_42 you mentioned your way to do so:

It would have been shorter to write f(x)=(x−α)g(x)∈C[x] and then discuss the neighborhood of α by continuity of g(x).

it is really short, indeed. But what if f(x) consists more than one time's (x-$\alpha$)?

 

[fresh_42]

it is really short, indeed. But what if $f(x)$ consists more than one time's $(x-\alpha )$?

Field extensions of fields with characteristic zero are separable, i.e. such a case cannot occur. However, as far as I could see, the proof doesn't change if we split $f(x)=(x-\alpha )^kg(x).$

 

[julian]

Hi @graphking. I figured case (b) was needed because the question asked for a $c > 0$ for all rational numbers $\not= \alpha$. In the combined cases (a) and (b) $p$ and $q$ are any integers you want them to be (as long as $p/q \not= \alpha$) and I was able to eliminate $p$ from the determination of such a $c$ (assuming $p/q \not= \alpha$).

 

[graphking]

I mean the inequation we needed to proof is irrelevant with p, so p can be variable(q is fitted/set/settled)and when it comes to the nearest case to $\alpha$, we know that is the case we need to proof for this certain q, because, with other p, the distance would be bigger, so the inequation would hold if the nearest case hold.

 

[graphking]

The Q7 can be used to find transcendental numbers. I see in Zorich's analysis, a exercise in chapter 2.2, the Liouville Theorem: Any irrational number can be "well approximated" by rational numbers is a transcendental number, the "well approximated" means:
we call an irrational number $\alpha$ is well approximated by rational numbers when for any natural numbers n, N, there exist a rational number p/q satisfiy: |$\alpha$-p/q|<1/(N*q^n).

 

[fresh_42]

If someone wants to complete the proof, i.e. find a proof where the choice of $c$ is independent from $p,q$ as required ($\forall \,\alpha \,\exists\, c \,\forall \,p,q$) instead of ($\forall \,\alpha \, \forall \,p,q\,\exists\, c$) which I think @julian has proven, you can either use my hint at the end of post #105 or investigate the problem along the lines $\alpha =a+ib \in \not\in \mathbb{R}$ with the MVT.

 

[julian]

So @Fred Wright got for problem 1:

\begin{align*}
I = \Gamma (\frac{1}{3}) \Gamma (\frac{2}{3}) \sum_{n=0}^\infty (-1)^n \frac{x^{3n}}{(3n)!}
\end{align*}

where $x = 3 \lambda$, which I'm getting as well. And he found that

\begin{align*}
\sum_{n=0}^\infty (-1)^n \frac{x^{3n}}{(3n)!} = \frac{2}{3} e^{\frac{x}{2}} \cos \left( \frac{\sqrt{3} x}{2} \right) + \frac{1}{3} e^{-x}
\end{align*}

@fresh_42 said the cosine function looks wrong, but I'm getting the same result as @Fred Wright.

@Fred Wright's answer can be simplified slightly by using the multiplication theorem:

\begin{align*}
\Gamma (z) \Gamma \left( z + \frac{1}{n} \right) \Gamma \left( z + \frac{2}{n} \right) \cdots \Gamma \left( z + \frac{n-1}{n} \right) = (2 \pi)^{(n-1)/2} n^{1/2 - nz} \Gamma (nz)
\end{align*}

From which we have

\begin{align*}
\Gamma \left( \frac{1}{3} \right) \Gamma \left( \frac{2}{3} \right)
&\ = \lim_{z \rightarrow 0} \Gamma \left( z + \frac{1}{3} \right) \Gamma \left( z + \frac{2}{3} \right)
\nonumber \\
& = (2 \pi) 3^{1/2} \lim_{z \rightarrow 0} \frac{\Gamma (3z)}{\Gamma (z)}
\nonumber \\
& = (2 \pi) 3^{1/2} \lim_{z \rightarrow 0} \frac{1}{3} \frac{3z \Gamma (3z)}{z \Gamma (z)}
\nonumber \\
& = 2 \pi \sqrt{3} \lim_{z \rightarrow 0} \frac{1}{3} \frac{\Gamma (3z + 1)}{\Gamma (z + 1)}
\nonumber \\
& = \frac{2 \pi}{\sqrt{3}} .
\end{align*}

So @Fred Wright's answer reads:

\begin{align*}
I = \frac{2 \pi}{3 \sqrt{3}} \left[ 2 e^{\frac{3 \lambda}{2}} \cos \left( \frac{3 \sqrt{3} \lambda}{2} \right) + e^{-3 \lambda} \right]
\end{align*}

 

[fresh_42]

So @Fred Wright got for problem 1:

\begin{align*}
I = \Gamma (\frac{1}{3}) \Gamma (\frac{2}{3}) \sum_{n=0}^\infty (-1)^n \frac{x^{3n}}{(3n)!}
\end{align*}

where $x = 3 \lambda$, which I'm getting as well. And he found that

\begin{align*}
\sum_{n=0}^\infty (-1)^n \frac{x^{3n}}{(3n)!} = \frac{2}{3} e^{\frac{x}{2}} \cos \left( \frac{\sqrt{3} x}{2} \right) + \frac{1}{3} e^{-x}
\end{align*}

@fresh_42 said the cosine function looks wrong, but I'm getting the same result as @Fred Wright.

@Fred Wright's answer can be simplified slightly by using the multiplication theorem:

\begin{align*}
\Gamma (z) \Gamma \left( z + \frac{1}{n} \right) \Gamma \left( z + \frac{2}{n} \right) \cdots \Gamma \left( z + \frac{n-1}{n} \right) = (2 \pi)^{(n-1)/2} n^{1/2 - nz} \Gamma (nz)
\end{align*}

From which we have

\begin{align*}
\Gamma \left( \frac{1}{3} \right) \Gamma \left( \frac{2}{3} \right)
&\ = \lim_{z \rightarrow 0} \Gamma \left( z + \frac{1}{3} \right) \Gamma \left( z + \frac{2}{3} \right)
\nonumber \\
& = (2 \pi) 3^{1/2} \lim_{z \rightarrow 0} \frac{\Gamma (3z)}{\Gamma (z)}
\nonumber \\
& = (2 \pi) 3^{1/2} \lim_{z \rightarrow 0} \frac{1}{3} \frac{3z \Gamma (3z)}{z \Gamma (z)}
\nonumber \\
& = 2 \pi \sqrt{3} \lim_{z \rightarrow 0} \frac{1}{3} \frac{\Gamma (3z + 1)}{\Gamma (z + 1)}
\nonumber \\
& = \frac{2 \pi}{\sqrt{3}} .
\end{align*}

So @Fred Wright's answer reads:

\begin{align*}
I = \frac{2 \pi}{3 \sqrt{3}} \left[ 2 e^{\frac{3 \lambda}{2}} \cos \left( \frac{3 \sqrt{3} \lambda}{2} \right) + e^{-3 \lambda} \right]
\end{align*}

My answer reads $\dfrac{2\pi}{\sqrt{3}}\cdot e^{-3\lambda }$.