Compute the integral $ \int_{Re(s)=2}\frac{x^{-s}}{s^3}ds\quad\text{for real}\ x>0. $

Question

Use the residue theorem to compute the integral $$ \int_{Re(s)=2}\frac{x^{-s}}{s^3}ds\quad\text{for real}\ x>0, $$ where the contour is oriented upwards. (Hint: treat the cases of $ x<1 $ and $ x\ge 1 $ separately.)

My Attempt:

Draw the semicircle $ \Gamma: 2+R\exp(i\theta), \theta\in[\frac \pi 2, \frac {3\pi}2], R>2 $. Then $$\begin{align} \int_{\Gamma}\frac{x^{-s}}{s^3}ds &=\int\limits_{\substack{Re(s)=2\\Im(s)=-R}}^{\substack{Re(s)=2\\Im(s)=R}}\frac{x^{-s}}{s^3}ds+\int_{\frac \pi2}^{\frac {3\pi}2}\frac{x^{-2-R\exp(i\theta)}}{(2+R\exp(i\theta))^3}R\exp(i\theta)id\theta. \end{align} $$ Now let $ R\to\infty $ and by the standard trick we know that the second term on the right tends to zero and the first term on the right is $ \int_{Re(s)=2}\frac{x^{-s}}{s^3}ds\quad\text{for real}\ x>0. $ By the residue theorem, $$ \begin{align} \int_{\Gamma}\frac{x^{-s}}{s^3}ds &=2\pi i Res(\frac{x^{-s}}{s^3}, 0)\\ &=2\pi i\frac{\ln^2 x}{2}\\ &=\pi i\ln^2 x .\end{align}$$ Therefore, $$ \int_{Re(s)=2}\frac{x^{-s}}{s^3}ds=\pi i\ln^2 x\quad\text{for real}\ x>0. $$

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Am I right? I am confused that the hint says that we should treat the cases of $ x<1 $ and $ x\ge 1 $ separately, but I didn't use this fact, why?

Answer 1

The half-plane where $x^{-s}$ decays depends on the sign of $\log x$.

• For $x \in (0,1)$ then $$\int_{\Re(s)=2} s^{-3}x^{-s}ds = \lim_{R \to \infty} \int_{\partial ([-R,2] +i[-R,R])} s^{-3}x^{-s}ds =2i\pi Res(s^{-3}x^{-s},s=0) = i\pi \log^2 x$$

• For $x \ge 1$ then $$\int_{\Re(s)=2} s^{-3}x^{-s}ds = -\lim_{R \to \infty} \int_{\partial ([2,R] +i[-R,R])} s^{-3}x^{-s}ds =0$$

Answer 2

Let $I(x)$ be given by the integral

$$\begin{align} I(x)&=\int_{\text{Re}(s)=2}\frac{x^{-s}}{s^3}\,ds\tag1 \end{align}$$

For $0<x<1$, $\log(x)<1$ and we evaluate the integral of $\frac{x^{-s}}{s^3}$ over the closed contour, $C_R$, comprised of $(i)$ the straight line segment from $2-iR$ to $2+iR$ and $(ii)$ the circular arc of radius $\sqrt{R^2+4}$ starting from $2+iR$ traversed counterclockwise and ending at $2-iR$. Proceeding, we have

$$\begin{align} \oint_{C_R}\frac{x^{-s}}{s^3}\,ds&=\int_{2-iR}^{2+iR}\frac{x^{-s}}{s^3}\,ds +\int_{\arctan(R/2)}^{3\pi/2+\arctan(2/R)}\frac{x^{-\sqrt{R^2+4}e^{i\theta}}}{(\sqrt{R^2+4}e^{i\theta})^3}\,i\sqrt{R^2+4}e^{i\theta}\,d\theta\tag2\\\\ &=2\pi i \text{Res}\left(\frac{x^{-s}}{s^3},s=0\right)\\\\ &=(2\pi i )\frac1{2}\left.\frac{d^2}{ds^2}\left(s^3\,\frac{x^{-s}}{s^3}\right)\right|_{s=0}\\\\ &=i\pi\log^2(x) \end{align}$$

Letting $R\to\infty$, the second integral on the right-hand side of $(2)$ approaches $0$ and we find for $0<x<1$

$$I(x)=i\pi\log^2(x)$$

Similarly, for $x>1$, we close the contour in the right-half plane. Inasmuch as no poles are enclosed, we find $I(x)=0$ for $x>1$.

Putting it together reveals

$$I(x)=\begin{cases}i\pi\log^2(x)&,0<x\le 1\\\\0&, x>1\end{cases}$$

Answer 3

Deleted comment as was unclear, so I will make it an answer; the main issue is that for $x>1$, the exponential term on the circular integral is high since $\cos(\theta) \leq 0$ on your domain, so the integral doesn't go to zero and the result is valid only for $x \leq 1$. For $x>1$ we can use the circular domain to the right so $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$ and same argument works but no residue, so answer zero, or just move the vertical $s$ line to the right towards $\infty$ and use rectangles and no residue again to get zero this way