Prove that $f_n(z)=\int_0^{n} e^{-zt^2}dt$ is analytic in $\Re e(z)>0$ for all n

Question

I want to prove that $f(z)=\int_0^{\infty} e^{-zt^2}dt$ is analytic in $\Re e(z)>0$.

If $f_{n}$ are analytics then f is analytics as $f_n$ converges uniformly to $f$.

Answer 1

In fact, $f_n$ is analytic on the whole $\Bbb C$; it's $f$, the limit, that is analytic on only a subset.

To see this, the most straightforward way to proceed is to slip $\dfrac {\Bbb d} {\Bbb d z}$ inside the integral (using Lebesgue's dominated convergence theorem, for instance), which gives

$$f_n ' (z) = -\int \limits _0 ^n \Bbb e ^{-z t^2} t^2 \ \Bbb d t ,$$

showing that $f_n$ is complex-differentiable, and thus analytic.

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Alternatively, writing $z = x + \Bbb i y$ and $f_n = u_n + \Bbb i v_n = \Bbb e ^{- x t^2} \cos (y t^2) - \Bbb i \Bbb e ^{- x t^2} \sin (y t^2)$, you may check that the Cauchy-Riemann relations are verified:

$$\frac {\partial u_n} {\partial x} = \frac {\partial} {\partial x} \int \limits _0 ^n \Bbb e ^{-x t^2} \cos (y t^2) \ \Bbb d t = \int \limits _0 ^n (-t^2) \Bbb e ^{-x t^2} \cos (y t^2) \ \Bbb d t \\ \frac {\partial v_n} {\partial y} = \frac {\partial} {\partial y} \int \limits _0 ^n - \Bbb e ^{-x t^2} \sin (y t^2) \ \Bbb d t = \int \limits _0 ^n - \Bbb e ^{-x t^2} \cos (y t^2) (t^2) \ \Bbb d t$$

so these are equal, and a similar computation for the other pair of derivatives.

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Another option would be to exhibit the Taylor series of $f_n$, centered in $0$:

$$f_n (z) = \int \limits _0 ^n \Bbb e ^{-z t^2} \ \Bbb d t = \int \limits _0 ^n \sum _{k \ge 0} \frac {(-1)^k} {k!} t^{2k} z^k \ \Bbb d t = \dots$$

Notice that the convergence of that series is uniform on $\Bbb C$, therefore we may interchange summation and integration, in order to continue:

$$\dots = \sum _{k \ge 0} \frac {(-1)^k} {k!} z^k \int \limits _0 ^n t^{2k} \ \Bbb d t = \sum _{k \ge 0} \frac {(-1)^k} {k!} \frac {n^{2k+1}} {2k+1} z^k ,$$

which shows that, indeed, $f_n$ is holomorphic (in fact, it is even entire).

Answer 2

The other answer shows why $f_n$ is analytic. To show that $f$ is analytic on $\operatorname{Re} z > 0$, it's enough to show that it's analytic on $U_\epsilon = \{z : \operatorname{Re} z > \epsilon\}$ for any $\epsilon>0$.

We do this by showing $f_n \to f$ uniformly on $U_\epsilon$. Then, since a uniform limit of analytic functions is analytic, $f$ will be analytic on $U_\epsilon$. So, if $\operatorname{Re}z > \epsilon$: $$ \left\lvert f(z) - f_n(z) \right\rvert = \left\lvert \int_n^\infty e^{-zt^2} dt \right\rvert \le \int_n^\infty \left\lvert e^{-zt^2} \right\rvert dt \le \int_n^\infty e^{-\epsilon t^2} dt. $$ Now argue that the right hand side converges to $0$ as $n \to \infty$.

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edit: Note that convergence isn't uniform on $\{z : \operatorname{Re}(z) > 0\}$ , so this trick is necessary. For instance: $$ f(1/n^2) - f_n(1/n^2) = n \int_1^\infty e^{-u^2} du \to \infty. $$

Answer 3

Apply Morera directly (together with Fubini)

Morera's Theorem:

A function $f$ defined on a connected open set is analytic if and only if $\int_\gamma f(z)\;dz = 0$ for all simple closed curves $\gamma$.

(In general, interchanging two integrals is easier than interchanging an integral and a derivative.)

** added **

Here's how I want to use Fubini... Let $\gamma$ be a closed curve in the right half-plane.

For fixed $t$, the function $e^{-zt^2}$ is an analytic function of $z$. So by Morera, $\int_\gamma e^{-zt^2}\;dz = 0$. Then apply Fubini: $$ \int_\gamma f(z)\;dz = \int_\gamma\int_0^\infty e^{-zt^2}\;dt\;dz = \int_0^\infty \int_\gamma e^{-zt^2}\;dz\;dt = \int_0^\infty 0\;dt = 0 $$ By Morera the other direction, $f$ is analytic.

Answer 4

(just to mention it)

Let $$f(z) = \int_{0}^{\infty} e^{-zt^2}dt\qquad\qquad\qquad \scriptstyle(\,\text{Re}(z)\, >\, 0\,)$$ With the change of variable $u = z^{1/2}t$ we have $\displaystyle f(z) = z^{-1/2}\int_{0}^{z^{1/2}\infty} e^{-u^2} du$.

Note that $e^{-t^2}$ is analytic and decreases exponentially as $\text{Re}(t^2)\to \infty$, which is the case when $t^2 = z u^2, u \to \infty, \text{Re}(z) > 0$. We can apply the Cauchy integral theorem to $\int_{0}^{z^{1/2}R}+\int_{z^{1/2}R}^{R}+\int_{R}^{0} e^{-u^2} du = 0$ and obtain as $R \to \infty$ : $$\displaystyle\int_{0}^{z^{1/2}\infty}e^{-u^2}du = \int_{0}^{\infty}e^{-u^2}du = \frac{1}{2}\int_{-\infty}^\infty e^{-t^2}dt = \frac{\sqrt{\pi}}{2}$$ and $$\boxed{\ \ f(z) = \frac{\sqrt{\pi}}{2}z^{-1/2} \qquad \scriptstyle(\,\text{Re}(z)\, >\, 0\,) \ \ }$$