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Use the residue theorem to compute the integral
$$ int_Re(s)=2fracx^-ss^3dsquadtextfor real x>0, $$
where the contour is oriented upwards. (Hint: treat the cases of $ x<1 $ and $ xge 1 $ separately.)

My Attempt:

Draw the semicircle $ Gamma: 2+Rexp(itheta), thetain[frac pi 2, frac 3pi2], R>2 $.
Then $$beginalign int_Gammafracx^-ss^3ds &=intlimits_substackRe(s)=2\Im(s)=-R^substackRe(s)=2\Im(s)=Rfracx^-ss^3ds+int_frac pi2^frac 3pi2fracx^-2-Rexp(itheta)(2+Rexp(itheta))^3Rexp(itheta)idtheta.
endalign
$$
Now let $ Rtoinfty $ 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)=2fracx^-ss^3dsquadtextfor real x>0. $ By the residue theorem,
$$ beginalign int_Gammafracx^-ss^3ds &=2pi i Res(fracx^-ss^3, 0)\
&=2pi ifracln^2 x2\
&=pi iln^2 x .endalign$$
Therefore, $$ int_Re(s)=2fracx^-ss^3ds=pi iln^2 xquadtextfor real x>0. $$

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

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

• For $x in (0,1)$ then $$int_Re(s)=2 s^-3x^-sds = lim_R to infty int_partial ([-R,2] +i[-R,R]) s^-3x^-sds =2ipi Res(s^-3x^-s,s=0) = ipi log^2 x$$




• For $x ge 1$ then $$int_Re(s)=2 s^-3x^-sds = -lim_R to infty int_partial ([2,R] +i[-R,R]) s^-3x^-sds =0$$

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 $-fracpi2 leq theta leq fracpi2$ 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

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

$$beginalign
I(x)&=int_textRe(s)=2fracx^-ss^3,dstag1
endalign$$

For $0<x<1$, $log(x)<1$ and we evaluate the integral of $fracx^-ss^3$ 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 $sqrtR^2+4$ starting from $2+iR$ traversed counterclockwise and ending at $2-iR$. Proceeding, we have

$$beginalign
oint_C_Rfracx^-ss^3,ds&=int_2-iR^2+iRfracx^-ss^3,ds +int_arctan(R/2)^3pi/2+arctan(2/R)fracx^-sqrtR^2+4e^itheta(sqrtR^2+4e^itheta)^3,isqrtR^2+4e^itheta,dthetatag2\\
&=2pi i textResleft(fracx^-ss^3,s=0right)\\
&=(2pi i )frac12left.fracd^2ds^2left(s^3,fracx^-ss^3right)right|_s=0\\
&=ipilog^2(x)
endalign$$

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

$$I(x)=ipilog^2(x)$$

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

$$I(x)=begincasesipilog^2(x)&,0<xle 1\\0&, x>1endcases$$