Question 10.35: (a) Let p > 0. Prove that F(z) = ∫z^∞ e^-t/t^p dt = e^-z ...
(a) Let p>0. Prove that
\begin{aligned} F(z)=\int\limits_{z}^{\infty} \frac{e^{-t}}{t^{p}} d t&=e^{-z}\left\{\frac{1}{z^{p}}-\frac{p}{z^{p+1}}+\frac{p(p+1)}{z^{p+2}}-\cdots(-1)^{n} \frac{p(p+1) \cdots(p+n-1)}{z^{p+n}}\right\} \\ & +(-1)^{n+1} p(p+1) \cdots(p+n) \int\limits_{z}^{\infty} \frac{e^{-t}}{t^{p+n+1}} d t \end{aligned}
(b) Use (a) to prove that
F(z)=\int\limits_{z}^{\infty} \frac{e^{-t}}{t^{p}} d t \sim e^{-z}\left\{\frac{1}{z^{p}}-\frac{p}{z^{p+1}}+\frac{p(p+1)}{z^{p+2}}-\cdots\right\}=S(z)
that is, the series on the right is an asymptotic expansion of the function on the left.
(a) Integrating by parts, we have
\begin{aligned} I_{p}=\int\limits_{z}^{\infty} \frac{e^{-t}}{t^{p}} d t & =\underset{M \rightarrow \infty}{\lim} \int\limits_{z}^{M} e^{-t} t^{-p} d t=\lim _{M \rightarrow \infty}\left\{\left.\left(-e^{-t}\right)\left(t^{-p}\right)\right|_{z} ^{M}-\int\limits_{z}^{M}\left(-e^{-t}\right)\left(-p t^{-p-1}\right) d t\right\} \\ & =\underset{M \rightarrow \infty}{\lim}\left\{\frac{e^{-z}}{z^{p}}-\frac{e^{-M}}{M^{p}}-p \int\limits_{z}^{M} \frac{e^{-t}}{t^{p+1}} d t\right\}=\frac{e^{-z}}{z^{p}}-p \int\limits_{z}^{\infty} \frac{e^{-t}}{t^{p+1}} d t=\frac{e^{-z}}{z^{p}}-p I_{p+1} \end{aligned}
Similarly, I_{p+1}=\left(e^{-z} / z^{p+1}\right)-(p+1) I_{p+2} so that
I_{p}=\frac{e^{-z}}{z^{p}}-p\left\{\frac{e^{-z}}{z^{p+1}}-(p+1) I_{p+2}\right\}=\frac{e^{-z}}{z^{p}}-\frac{p e^{-z}}{z^{p+1}}+p(p+1) I_{p+2}
By continuing in this manner, the result follows.
(b) Let
S_{n}(z)=e^{-z}\left\{\frac{1}{z^{p}}-\frac{p}{z^{p+1}}+\frac{p(p+1)}{z^{p+2}}-\cdots(-1)^{n} \frac{p(p+1) \cdots(p+n-1)}{z^{p+n}}\right\}
Then
R_{n}(z)=F(z)-S_{n}(z)=(-1)^{n+1} p(p+1) \cdots(p+n) \int\limits_{z}^{\infty} \frac{e^{-t}}{t^{p+n+1}} d t
Now, for real z>0,
\begin{aligned}\left|R_{n}(z)\right|=p(p+1) \cdots(p+n) \int\limits_{z}^{\infty} \frac{e^{-t}}{t^{p+n+1}} d t & \leq p(p+1) \cdots(p+n) \int\limits_{z}^{\infty} \frac{e^{-t}}{z^{p+n+1}} d t \\& \leq \frac{p(p+1) \cdots(p+n)}{z^{p+n+1}}\end{aligned}
since
\int\limits_{z}^{\infty} e^{-t} d t \leq \int\limits_{0}^{\infty} e^{-t} d t=1
Thus
\underset{z \rightarrow \infty}{\lim}\left|z^{n} R_{n}(z)\right| \leq \underset{z \rightarrow \infty}{\lim} \frac{p(p+1) \cdots(p+n)}{z^{p}}=0
and it follows that \lim _{z \rightarrow \infty} z^{n} R_{n}(z)=0. Hence, the required result is proved for real z>0. The result can also be extended to complex values of z.
Note that since
\left|\frac{u_{n+1}}{u_{n}}\right|=\left|\frac{p(p+1) \cdots(p+n) / z^{p+n+1}}{p(p+1) \cdots(p+n-1) / z^{p+n}}\right|=\frac{p+n}{|z|}
where u_{n} is the nth term of the series, we have for all fixed z
\underset{n \rightarrow \infty}{\lim}\left|\frac{u_{n+1}}{u_{n}}\right|=\infty
and the series diverges for all z by the ratio test.