For each fixed $n$, show that $$f_n(z)=\int_1^nt^{z-1}e^{-t}dt$$ is an entire function of $z$.

From Morera 's theorem:

If a continuous, complex-valued function $f$ in a domain $D$ that satisfies

$$\oint_\gamma f(z)\,dz = 0 $$

for every closed contour in $D$, then $f$ is analytic.

From this theorem how can I show that my function is indeed continuous? Do I prove this straight from the definition of continuity?

A hint which was provided, which I don't understand why, goes:$$\text{Look at:} \ |f(z_1)-f(z_2)|\le \int_1^n\frac{e^{-t}}{t}|t^{z_1}-t^{z_2}|dt \ \text{where},\\ z_1=x_1+iy_1 \\ z_2=x_2+iy_2$$

thus I must bound this but I don't know how to go from here? Any hints please?

  • $\begingroup$ You don't need Morera if you only care about continuity. THe hint suggests that you bound $|t^{z_1}-t^{z_2}|$ by $M|z_1-z_2|$ uniformly on $[1,n]$. This will show after integration that $f$ is Lipschitz, hence continuous. $\endgroup$ – Julien Apr 16 '13 at 5:13
  • $\begingroup$ Here is how you can use Morera's theorem. $\endgroup$ – Mhenni Benghorbal Apr 16 '13 at 5:16
  • $\begingroup$ @julien I am having trouble continuing so that it is bounded to $M|z_1-z_2|$. $\endgroup$ – Tom Apr 16 '13 at 5:18
  • $\begingroup$ @MhenniBenghorbal Yes I have looked at that before posting this question but I still don't understand how to go further along with the hint given using that. $\endgroup$ – Tom Apr 16 '13 at 5:19
  • $\begingroup$ @julien I am mistaken, I meant to say entire not just continuous. $\endgroup$ – Tom Apr 16 '13 at 5:20

Related problems: I. Just see this,

$$ \oint_\gamma f_n(z) dz = \oint_\gamma \int_1^nt^{z-1}e^{-t}dt dz = \int_1^n e^{-t}\frac{dt}{t}\oint_\gamma e^{z\ln t} dz \,dt = \dots. $$

Can you finish it. What kind of function is $e^{z\ln(t)}$?

Added: To prove continuity, advance like that

$$ |f_n(z+h)-f_n(z)| = \Big|\int_1^n (e^{(z+h)\ln t}-e^{ z \ln t }) \frac{e^{-t}}{t}dt \Big| $$

$$ \leq \int_1^n \big|e^{(z+h)\ln t}-e^{ z \ln t }\big| \frac{e^{-t}}{t}dt \leq M e^{-1} |h|\int_{1}^{n} \ln t\frac{1}{t}dt\dots. $$

Can you finish the task? Notice that, I used the mean value theorem to find the last inequality and $ e^{-t} \leq e^{-1} $ for $t\in [1,n]$.

  • $\begingroup$ Why would $e^{z\ln(t)}$ complete this proof? $\endgroup$ – Tom Apr 16 '13 at 5:38
  • $\begingroup$ @user71642: Just note that, $e^{z\ln(t)}$ is an analytic function in $z$, so the inner integral is zero by Cauchy theorem, That implies $\oint_\gamma f_n(z) dz =0$. $\endgroup$ – Mhenni Benghorbal Apr 16 '13 at 5:46
  • $\begingroup$ Thank you! And how will I show that it is continuous? If it is continuous and analytic then it is entire which completes the proof? $\endgroup$ – Tom Apr 16 '13 at 5:50
  • $\begingroup$ @user71642: You are very welcome. $\endgroup$ – Mhenni Benghorbal Apr 16 '13 at 6:20

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