Mellin Transform of Cosine Analytic Continuation

9.5 Analytic continuation of Mellin transforms

Now $$M\left[e^{-x};s\right]={\int \; <!--nolimits-->}_0^{\infty }{x}^{s-1}{e}^{-x}dx=\Gamma \left(s\right)$$

is analytic in the right-hand plane $0<\Re \left(s\right).$

Consider next the MT of the function $e^{-it}$. This is given by $$M\left[e^{-ix};s\right]={\int \; <!--nolimits-->}_0^{\infty }{x}^{s-1}{e}^{-ix}dx.$$

Note that the integral does not converge absolutely, only conditionally. We can evaluate the integral by considering $${\oint }_C{z}^{s-1}{e}^{-iz}dz$$

where the contour $C$ is as shown in figure 10.1 .

---

Figure 9.16: Contour $C=L_1+C_1+L_2+C_2$

---

From Cauchy's theorem $${\oint }_C{z}^{s-1}{e}^{-iz}dz=0$$

since the integrand is analytic inside $C.$ Now $$\left|I_{C_2}\right|=\left|{\int \; <!--nolimits-->}_{-\frac{\pi }{2}}^{0}i\delta e^{i\theta }{\left(\delta e^{i\theta }\right)}^{s-1}{e}^{-i\delta \left(cos\theta +isin\theta \right)}d\theta \right|\to 0,$$

as $\delta \to 0$ provided $\Re \left(s\right)>0.$ Next with $s=\sigma +i\lambda$ $$\left|{\int \; <!--nolimits-->}_{C_1}\right|=\left|-{\int \; <!--nolimits-->}_{-\frac{\pi }{2}}^0{\left(R{e}^{i\theta }\right)}^{s-1}iR{e}^{i\theta }{e}^{-iRcos\theta +Rsin\theta }d\theta \right|.$$

So $$\left|{\int \; <!--nolimits-->}_{C_1}\right|\le \left|e^{\left|\lambda \right|\pi ∕2}{\int \; <!--nolimits-->}_0^{\frac{\pi }{2}}{e}^{\lambda \theta }{R}^{\sigma }{e}^{-Rsin\theta }d\theta \right|$$

$$\le e^{\left|\lambda \right|\pi ∕2}{R}^{\sigma }{e}^{\left|\lambda \right|\frac{\pi }{2}}{\int \; <!--nolimits-->}_0^{\frac{\pi }{2}}{e}^{-\frac{2R\theta }{\pi }}d\theta$$

$$<\pi e^{\left|\lambda \right|\pi ∕2}{e}^{\left|\lambda \right|\frac{\pi }{2}}{R}^{\sigma -1}.$$

Thus $$\left|{\int \; <!--nolimits-->}_{C_1}\right|\to 0\text{as}R\to \infty$$

provided $\Re \left(s\right)<1.$ Hence taking the limit as $\delta \to 0$ and $R\to \infty$ we obtain $${\int \; <!--nolimits-->}_{L_1}+{\int \; <!--nolimits-->}_{L_2}=0$$

giving $${\int \; <!--nolimits-->}_0^{\infty }{x}^{s-1}{e}^{-ix}dx={\int \; <!--nolimits-->}_0^{\infty }{\left(e^{-i\frac{\pi }{2}}y\right)}^{s-1}{e}^{-i\frac{\pi }{2}}{e}^{-y}dy$$

$$=e^{-is\frac{\pi }{2}}{\int \; <!--nolimits-->}_0^{\infty }{y}^{s-1}{e}^{-y}dy=e^{-is\frac{\pi }{2}}\Gamma \left(s\right)$$

Hence the Mellin transform of $e^{ix}$ is given by

$$M\left[e^{-ix};s\right]=\Gamma \left(s\right)e^{-is\frac{\pi }{2}},$$

(9.5.1)

provided $0<\Re \left(s\right)<1.$

We can analytically continue $M\left[e^{-ix};s\right]$ into the right half-plane $\Re \left(s\right)\ge 1$ using the formula (9.5.1 ).

---

---

Similarly we showed earlier that the function $f\left(x\right)=1∕\left(1+x\right)$ has the Mellin transform

$$M\left[\frac{1}{1+x};s\right]=\frac{\pi }{sin\left(\pi s\right)},$$

(9.5.2)

again provided $0<\Re \left(s\right)<1.$ We can use (9.5.2 ) to analytically continue $M\left[1∕\left(1+x\right);s\right]$ into the right half-plane $\Re \left(s\right)\ge 1$. The analytic continuation gives a function which has simple poles at the positive integers.

---

---

Let $F\left(s\right)=M\left[f\left(x\right);s\right]$ be the Mellin transform of $f\left(x\right)$ which is analytic in the strip $\alpha <\Re \left(s\right)<\beta$. Suppose that as $x\to \infty$ we have $$f\left(x\right)~e^{-d{x}^{\nu }}\sum _{m=0}^{\infty }{x}^{-r_m}\sum _{n=0}^{N\left(m\right)}{c}_{mn}{\left(logx\right)}^n,$$

where $\Re \left(d\right)\ge 0,\nu >0$ and $\Re \left(r_m\right)$ is an increasing sequence in $m$, and $N\left(m\right)$ are non-negative integers. Then $F\left(s\right)$ can be continued analytically into the right-half plane $\beta <\Re \left(s\right)$. The analytic continuation is such that

1. If $\Re \left(d\right)>0$, then $F\left(s\right)$ is analytic in the right-half plane.
2. If $d=-i\omega ,\omega \ne 0$ then $F\left(z\right)$ is analytic in the right half-plane.
3. If $d=0$ then $F\left(s\right)$ is analytic in the right-half plane except for poles.

The proofs of the results are given in detail in (Bleistein & Handelsman , 1975, chap. 4). Outline proofs are as follows:

Proof of (1) This follows from the properties of the Mellin transform given earlier. Proof of (2) Write $$F\left(s\right)={\int \; <!--nolimits-->}_0^{\infty }{x}^{s-1}f\left(x\right)dx={\int \; <!--nolimits-->}_0^1{x}^{s-1}f\left(x\right)dx+{\int \; <!--nolimits-->}_1^{\infty }{x}^{s-1}f\left(x\right),dx,$$

and express the second integral as $${\int \; <!--nolimits-->}_1^{\infty }\left(f\left(x\right)-h_k\left(x\right)\right)x^{s-1}dx+{\int \; <!--nolimits-->}_1^{\infty }{h}_k\left(x\right)x^{s-1}dx,$$

where $$h_k\left(x\right)=e^{i\omega x^{\nu }}\sum _{\Re \left(r_m\right)<k}{x}^{-r_m}\sum _{n=0}^{N\left(m\right)}{c}_{mn}{\left(logx\right)}^n.$$

Note that $f\left(x\right)-h_k\left(x\right)=O\left(e^{i\omega x^{\nu }}{x}^{-\Re \left(r_j\right)}{\left(logx\right)}^{N\left(j\right)}\right)$ where $j$ is such that $\Re \left(r_j\right)\ge k$.

So the first term is analytic in the extended strip $\alpha <\Re \left(s\right)<k.$

We need to prove that the term $${\int \; <!--nolimits-->}_1^{\infty }{h}_k\left(x\right)x^{s-1}dx,$$

can be analytically continued in the extended strip.

Let $h_k\left(x\right)=e^{i\omega x^{\nu }}{x}^{-r_0}{H}_k\left(x\right),$ where $H_k\left(x\right)=O\left({\left(logx\right)}^{N\left(0\right)}\right)$ as $x\to \infty .$

Now $${\int \; <!--nolimits-->}_1^{\infty }{h}_k\left(x\right)x^{s-1}dx,={\int \; <!--nolimits-->}_1^{\infty }{e}^{i\omega x^{\nu }}{x}^{s-r_0-1}{H}_k\left(x\right)dx$$

$$={\int \; <!--nolimits-->}_1^{\infty }{e}^{i\omega x^{\nu }}{x}^{\nu -1}{x}^{s-r_0-\nu }{H}_k\left(x\right)dx$$

$$={\left[\frac{x^{s-r_0-\nu }{e}^{i\omega x^{\nu }}{H}_k\left(x\right)}{i\omega \nu }\right]}_1^{\infty }$$

$$-{\int \; <!--nolimits-->}_1^{\infty }{e}^{i\omega x^{\nu }}{x}^{s-r_0-\nu -1}\left[\frac{\left(s-r_0-\nu \right)x{H}_k\left(x\right)+x\frac{d{H}_k\left(x\right)}{dx}}{i\omega \nu }\right]dx$$

The last term is of the form $${\int \; <!--nolimits-->}_1^{\infty }{x}^{s-r_0-\nu -1}{e}^{i\omega x^{\nu }}{\tilde{H}}_k\left(x\right)dx,$$

and is analytic in the extended region $\Re \left(z\right)<\Re \left(r_0\right)+\nu .$

By integrating by parts $q$ times the integral $${\int \; <!--nolimits-->}_1^{\infty }{h}_k\left(x\right)x^{s-1}dx,$$

can be analytically continued into the extended region $\alpha <\Re \left(s\right)<\Re \left(r_0\right)+q\nu .$

Putting it all together this shows that the Mellin transform of $f\left(x\right)$ can be analytically continued into the right-half plane $\alpha <\Re \left(s\right),$ as an analytic function.

Proof of (3) Introduce $$h_k\left(x\right)=\sum _{\Re \left(r_m\right)<k}{x}^{-r_m}\sum _{n=0}^{N\left(m\right)}{c}_{mn}{\left(logx\right)}^n.$$

and note that the Mellin transform of $f\left(x\right)$ is $${\int \; <!--nolimits-->}_0^{\infty }{x}^{s-1}f\left(x\right)dx$$

$$={\int \; <!--nolimits-->}_0^{1}f\left(x\right)x^{s-1}dx+{\int \; <!--nolimits-->}_1^{\infty }\left(f\left(x\right)-h_k\left(x\right)\right)dx+{\int \; <!--nolimits-->}_1^{\infty }{x}^{s-1}{h}_k\left(x\right)dx.$$

The first term is analytic in $\alpha <\Re \left(s\right).$

As before the term $$I_2={\int \; <!--nolimits-->}_1^{\infty }{x}^{s-1}\left(f\left(x\right)-h_k\left(x\right)\right)dx$$

is analytic in the strip $\alpha <\Re \left(s\right)<k.$

The last term can be computed directly noting that $${\int \; <!--nolimits-->}_1^{\infty }{x}^{s-r_m-1}{\left(logx\right)}^{n}dx=\frac{n!{\left(-1\right)}^{n+1}}{{\left(s-r_m\right)}^{n+1}}.$$

Hence $$M\left[f;s\right]=\sum _{\begin{array}{c}\hfill m\hfill \\ \hfill \Re \left(r_m\right)<k\hfill \end{array}}\sum _{n=0}^{N\left(m\right)}\frac{{\left(-1\right)}^{n+1}{c}_{mn}n!}{{\left(s-r_m\right)}^{n+1}}+I_2.$$

The right hand is analytic except for poles at $s=r_m$ in the extended strip $\alpha <\Re \left(s\right)<k$, and since $k$ is arbitrary this proves the result.

The analytic continuation of $M\left[f;s\right]$ into the left-half plane $\Re \left(s\right)<\alpha$ can be obtained and depends on the properties of $f\left(x\right)$ as $x\to 0+$

Suppose that as $x\to 0+$ we have $$f\left(x\right)~e^{-q{x}^{\mu }}\sum _{m=0}^{\infty }{x}^{a_m}\sum _{n=0}^{K\left(m\right)}{b}_{mn}{\left(logx\right)}^n,$$

where $\Re \left(q\right)\ge 0,\mu <0$ and $\Re \left(a_m\right)$ is an increasing sequence in $m$, and $K\left(m\right)$ are non-negative integers. Then $F\left(s\right)$ can be continued analytically into the left-half plane $\Re \left(s\right)<\alpha$. The analytic continuation is such that

1. If $\Re \left(q\right)>0$, then $F\left(s\right)$ is analytic in the left-half plane $\Re \left(s\right)<\beta$.
2. If $q=-i\omega ,\omega \ne 0$ then $F\left(z\right)$ is analytic in the left half-plane $\Re \left(s\right)<\alpha$.
3. If $q=0$ then $F\left(s\right)$ is analytic in the left-half plane except for poles at $s=-a_m$ with Laurent expansion $$\sum _{n=0}^{K\left(m\right)}\frac{b_{mn}{\left(-1\right)}^{m}n!}{{\left(s+a_m\right)}^{n+1}}.$$

In summary, the above results show that the Mellin transform of a function $f\left(x\right)$ , which has a particular type of asymptotic behaviour for small or large $x$, can be analytically continued into the whole complex plane with at worst pole singularities at points which can be predicted by the asymptotic behaviour of $f\left(x\right)$ for $x\to 0+$ and $x\to \infty .$

Example Consider $$f\left(x\right)=e^{ix}=\sum _{n=0}^{\infty }\frac{{\left(ix\right)}^n}{n!}.$$

Now

$$M\left[e^{ix};s\right]=\Gamma \left(s\right)e^{is\frac{\pi }{2}},$$

(9.5.3)

which is analytic in the strip $0<\Re \left(s\right)<1.$ The analytic continuation of the transform into the left-half plane using (9.5.3 ) shows that the function is analytic in $\Re \left(s\right)<0$ except for simple poles at $s=0,1,-2,\cdots -n,...,$ as predicted by the theorem.

Example

Consider $$Ei\left(x\right)={\int \; <!--nolimits-->}_x^{\infty }\frac{e^{-u}}{u}du$$

$$=-log\left(x\right)-\gamma -\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n{x}^n}{nn!}.$$

The Mellin transform of $Ei\left(x\right)$ is given by

$$M\left[Ei\left(x\right);s\right]=\frac{\Gamma \left(s\right)}{s},\Re \left(s\right)>0.$$

(9.5.4)

We can analytically continue (9.5.4 ) into the left-half plane and the analytic continuation gives a function which has simple poles at the negative integers and a double pole at $s=0.$

---

---

stampcancest2002.blogspot.com

Source: http://oldwww.ma.man.ac.uk/~gajjar/magicalbooks/methods/notest1_2se39.html

Share this post