11.4 The goal of this problem is to show that the constant $C$ that appears in $(11.25)$ well-defined and positive, i.e. that the integral $I_{0}=\int_{-\infty}^{\infty} M_{0}(\theta) d \theta$ exists and negative.
(a) Establish the divergence of $M_{0}(\theta)$ for large $|\theta|$. Namely, using $(11.23)$ deduce tl asymptotic
$$
M_{0} \rightarrow(3 \theta)^{1 / 3}-(3 \theta)^{-4 / 3} \quad \text { as }|\theta| \rightarrow \infty
$$
(b) Establish the validity of the relation $\int_{-\infty}^{\infty}(3 \theta)^{1 / 3} d \theta=0 .$ (The integra $\int_{-\infty}^{\infty}(\ldots) d \theta$ are defined as $\left.\lim _{\Theta \rightarrow \infty} \int_{-\Theta}^{\Theta}(\ldots) d \theta .\right)$
(c) Using the above relation rewrite $I_{0}$ as the integral $I_{0}=\int_{-\infty}^{\infty} w(\theta) d \theta$ with $w(\theta)$ $M_{0}(\theta)-(3 \theta)^{1 / 3} .$ Show that the asymptotic behavior of the function $w(\theta)$ impli that it is integrable, and hence the integral $I_{0}$ exists.
(d) To establish that the integral $I_{0}$ is negative it suffices to show that $w(\theta)$ is negativ This fact can be derived as follows:
(i) Recall that $w(\theta)$ is negative for $\theta \rightarrow-\infty$.
(ii) Using this previous assertion show that, if $w(\theta)$ is somewhere positive, the would be a first point $\vartheta$ where $w(\vartheta)=0 .$ Argue that the derivative $w^{\prime}(i$ at this point should be non-negative, since the graph of $w(\theta)$ crosses $w=$ from below at point $\vartheta$
(iii) Show by direct computation that
$$
\frac{d w}{d \theta}=\frac{d M_{0}}{d \theta}-(3 \theta)^{-2 / 3}=\theta-\frac{1}{3} M_{0}^{3}-(3 \theta)^{-2 / 3}
$$
Using this result, in conjunction with the presumed relation $M_{0}(\vartheta)$ $(3 \vartheta)^{1 / 3}$, deduce that the derivative at point $\vartheta$ is $w^{\prime}(\vartheta)=-(3 \vartheta)^{-2 / 3}$, i.e. is negative. This contradiction proves that $w$ is indeed negative.