### Author Topic: Q5 TUT 0202  (Read 2641 times)

#### Victor Ivrii

• Elder Member
• Posts: 2574
• Karma: 0
##### Q5 TUT 0202
« on: November 02, 2018, 03:30:30 PM »
$\newcommand{\Log}{\operatorname{Log}}$

Use Morera's Theorem to show that the following function is analytic on the indicated domain; find a power-series expansion for the function by using the known power series for the integrand and interchanging the summation and integration.
$$\int_0^{1/2} \Log (1-tz)\,dt\qquad\text{on}\; |z|<2.$$
« Last Edit: November 04, 2018, 09:44:04 PM by Victor Ivrii »

#### ZhenDi Pan

• Jr. Member
• Posts: 10
• Karma: 20
##### Re: Q5 TUT 0202
« Reply #1 on: November 02, 2018, 05:06:37 PM »
We have

\int_{0}^{1/2} \log(1 - t z) dt

Integrate over $\gamma$ with respect to $z$, consider

f(z)=\log(1-tz)

Function $f(z)$ is analytic on $\mid z \mid <2$, by Cauchy's theorem, for any closed curve $\gamma$

\int_\gamma f(z)dz = 0 \\
\int_{0}^{1/2} (\int_\gamma \log(1-zt) \,dz)\,dt = \int_{0}^{1/2} 0\,dt

So it is analytic on domain $\mid z\mid < 2$.

Since $\log(1 - t z) = \sum_{n=1}^\infty \frac{-( z t)^n}{n}$ is valid when $\mid zt \mid<1$, and since $\mid z \mid<2$, for all  $t \in [0,\frac{1}{2}]$. We have

\int_{0}^{1/2} \log(1 - t z) dt  = - \int_{0}^{1/2} \sum_{n=1}^\infty  \frac{( z t)^n}{n} dt \\
= - \sum_{n=1}^\infty  \int_{0}^{1/2}  \frac{( z t)^n}{n} dt \\
=  -\sum_{n=1}^\infty\frac{1}{2^{n+1} n (n+1)} z^n
« Last Edit: November 02, 2018, 06:18:56 PM by ZhenDi Pan »

#### Victor Ivrii

It is $\Log$ rather than $\log$ ; otherwise it would be multivalued