= Solution
For the <simple symmetric random walk>, the <moment-generating function> is
$$
M(\tau)=\frac{e^\tau+e^{-\tau}}2=\cosh\tau.
$$
For a given $\alpha>0$, choose the unique $\tau>0$ satisfying $\cosh\tau=e^\alpha$. The transform in part (d) becomes $\mathbb E e^{-\alpha T}=e^{-b\tau}$. To express this in terms of $\alpha$, put $r=e^{-\tau}\in(0,1)$. The equation $(r+r^{-1})/2=e^\alpha$ gives $r^2-2e^\alpha r+1=0$. Its smaller root is the one in $(0,1)$, so
$$
\boxed{\mathbb E[e^{-\alpha T}]
=\left(e^\alpha-\sqrt{e^{2\alpha}-1}\right)^b
=\left(\frac{e^{-\alpha}}{1+\sqrt{1-e^{-2\alpha}}}\right)^b.}
$$
The second expression follows by rationalizing the first root. As $\alpha\downarrow0$ the transform tends to one, consistent with the almost-sure finiteness proved in part (c).
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