I want to show that some integration with vertical line is bounded.
function $f(\mu)$ is given by $$ f(\mu)=A^{-\sqrt{\mu}} \frac{(B_1-\sqrt\mu)}{(B_2-\sqrt\mu)(B_3+\sqrt\mu)} $$ where $f$ is defined on some half space $\{a+bi; a>a_0\}$ and $A>1$, $B_1, B_2, B_3>0$. Here, show that $$ \int_{-\infty}^{\infty} |f(x+yi)|^2 dy<M $$ for some real value $M$.
Could you explain the process line by line? Thanks in advance!.
P.S. I am doing this because of the Paley-Weiener theorem.
On
Let us split the integral into $\int \limits _{- \infty} ^{-R} + \int \limits _{-R} ^R + \int \limits _R ^\infty$, where $R>1$ is large enough in order for the possible roots of the denominator to be included in $[-R, R]$. Let us first analyze the third integral, for $y > R$, which is a type 1 improper integral.
Let us show that this integral is convergent using techniques for improper integrals. Let us introduce the notation: $f \sim g$ if and only if $\lim \limits _{t \to \infty} \frac {f(t)} {g(t)} \in (1, \infty)$. By the limit comparison test, if $f$ and $g$ are bounded, integrable on $[R, R'] \; \forall R' > R$ and $f \sim g$ then $\int \limits _R ^\infty f(t) \Bbb d t$ and $\int \limits _R ^\infty f(t) \Bbb d t$ have the same nature. We already have $f$, we shall construct an integrable $g$ to which we shall apply the above test.
Note that the integrand can be successively simplified by taking $\lim \limits _{y \to \infty}$ as follows:
$$\Big| A^{- \sqrt y \sqrt {\frac x y + \Bbb i}} \frac {\sqrt y (\frac {B_1} {\sqrt y} - \sqrt {\frac x y + \Bbb i} )} {\sqrt y ( \frac {B_2} {\sqrt y}- \sqrt {\frac x y + \Bbb i} ) \sqrt y ( \frac {B_3} {\sqrt y} + \sqrt {\frac x y + \Bbb i} )} \Big| \sim \\ \Big| A^{- \sqrt y \sqrt {\frac x y + \Bbb i}} \frac 1 {\sqrt y} \frac {- \sqrt {\Bbb i}} {-\sqrt {\Bbb i} \sqrt {\Bbb i}} \Big| = \\ \Big| A^{- \sqrt y \sqrt {\frac x y + \Bbb i}} \Big| \frac 1 {\sqrt y} = \\ \Big| A^{- \sqrt y \sqrt {\frac x y + \Bbb i} + \sqrt y \sqrt {\Bbb i}} \; A^{- \sqrt y \sqrt {\Bbb i}} \Big| \frac 1 {\sqrt y} = \\ \Big| A^{\sqrt y \frac {- \frac x y} {\sqrt {\Bbb i} + \sqrt {\frac x y + \Bbb i}}} A^{- \sqrt y \sqrt {\Bbb i}} \Big| \frac 1 {\sqrt y} \sim \\ \Big| A^{- \sqrt y \sqrt {\Bbb i}} \Big| \frac 1 {\sqrt y} = \\ \frac {A^{- \frac {\sqrt y} {\sqrt 2}}} {\sqrt y} ,$$
and this will be our $g$. It is obviously bounded. To show that it is integrable on $[R, \infty)$, make the substitution $u = \frac {\sqrt y} {\sqrt 2}$ to obtain $\int \limits _{\sqrt {\frac R 2}} ^\infty \frac {A^{-u}} {\sqrt 2 u} 4 u \; \Bbb d u $ which is easily integrable and has a finite value.
An almost identical type of reasoning is used to show that the first integral is convergent too, using $\sqrt {-y}$ where we previously used $\sqrt y$.
Finally, let us investigate the middle integral. The only possible root of the denominator can only be produced by the factor $B_2 - \sqrt \mu$, giving $x = B_2 ^2, y=0$. Therefore, if $x \ne B_2 ^2$, the integrand in the middle integral is continuous on a compact interval, therefore has a finite integral, so the whole big integral is convergent.
If, on the other side, $x = B_2 ^2$, then the middle integral will be a type 2 improper integral, having a singularity in $B_2 ^2$. If $B_1 = B_2$ then the two quantities between parantheses will simplify and the middle integral will again be finite. If $B_1 \ne B_2$ then by the limit comparison test the middle integrand will behave like $A^{-\sqrt {B_2}} \frac {B_1 - B_2} {B_3 + B_2} \frac 1 {B_2 - \sqrt \mu}$ (replace $\mu$ by $B_2 ^2$ where possible). Therefore, we must investigate
$$\int \limits _{-R} ^R \frac 1 {B_2 - \sqrt {B_2 ^2 + \Bbb i y}} \Bbb d y = \int \limits _{-R} ^R \frac {B_2 + \sqrt {B_2 ^2 + \Bbb i y}} {- \Bbb i y} \Bbb d y .$$
Note that towards $y=0$ the integrand behaves like $\frac 1 y$ which is clearly not integrable. Therefore, in this case the middle integral is divergent, and so will be the whole integral.
To summarize:
In order to make $f$ well defined we have to assume $a_0>0$ and $B_2<\sqrt{a_0}$. Denote the domain ${\rm Re}(z)>a_0$ by $\Omega$. Let $z=a+it\in\Omega$. Then by a standard formula for square roots of complex numbers one has $${\rm Re}\bigl(\sqrt{z}\bigr)=\sqrt{{1\over2}\bigl(\sqrt{a^2+t^2}+a\bigr)}\geq \sqrt{a_0}\ .\tag{1}$$ It follows that $$|B_2-\sqrt{z}|\geq{\rm Re}\bigl(\sqrt{z}-B_2\bigr)\geq \sqrt{a_0}-B_2$$ and similarly $$|B_3+\sqrt{z}|\geq{\rm Re}\bigl(\sqrt{z}+B_3\bigr)\geq \sqrt{a_0}+B_3$$ for all $z\in\Omega$. These estimates prove that the expression $$g(z):={B_1-\sqrt{z}\over(B_2-\sqrt{z})(B_3+\sqrt{z})}={p\over B_2-\sqrt{z}}+{q\over B_3+\sqrt{z}}\ ,$$ where $p$ and $q$ are constants depending on the data $a_0$, $B_1$, $B_2$, $B_3$, is bounded on $\Omega$. On the other hand from $(1)$ we can also conclude that $${\rm Re}\bigl(\sqrt{a+it}\bigr)\geq\sqrt{{|t|\over2}}\ .$$ It follows that $$\left|A^{-\sqrt{z}}\right|=A^{-{\rm Re}\bigl(\sqrt{z}\bigr)}\leq A^{-\sqrt{t/2}}\ .$$ In all, we now have $$\bigl|f(a+it)\bigr|\leq C A^{-\sqrt{t/2}}\qquad(a\geq a_0, \ t\in{\mathbb R})\ ,$$ and it is easy to verify that $$\int_{-\infty}^\infty \bigl|f(a+it)\bigr|^2\>dt\leq M$$ for all $a\geq a_0$ and a suitable $M$.