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This is a complex analysis question, I tried to find the residue of the integrand at x=0, which equal to $x^p=0$. So it looks the integral naively equal to $0$. What did I do wrong?

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    $\begingroup$ If $p\in(0,1)$, the origin is a branch point for $\frac{x^{p-1}}{1+x}$. Which contour did you choose in order to apply the residue theorem? The simplest approach in my opinion is a real-analytic one, i.e. to enforce the substitution $\frac{1}{1+x}$, the properties of Euler's Beta function and the reflection formula for the $\Gamma$ function. $\endgroup$ – Jack D'Aurizio Nov 1 '17 at 23:20
  • $\begingroup$ Also: naively, the integral of a continuous positive function is positive. $\endgroup$ – Jack D'Aurizio Nov 1 '17 at 23:22
  • $\begingroup$ I chose a semi circle contour from 0 to $\pi/2$ with radius 0 to $\infty$. Also, I am not familiar with the Beta function and reflection formula for the gamma function. (I am an undergrad physicist :p) $\endgroup$ – Henry Wang Nov 1 '17 at 23:27
  • $\begingroup$ @HenryWang what is the complex function you are considering in order to evaluate this integral? Remember we always expect this result to pop out of evaluating the complex analog, it may not be that looking at $f(z)= \frac{x^{p-1}}{1+z}$ is the best choice. Further, note that if you DO decide to choose that function, the residue of the FUNCTION (we talk about residues of functions, not integrals) is at $z=-1$ on the real axis. $\endgroup$ – DaveNine Nov 1 '17 at 23:33
  • $\begingroup$ @HenryWang There is the Bromwich contour you always wished i.stack.imgur.com/aOopD.png $\endgroup$ – reuns Nov 1 '17 at 23:36
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basicly it comes from this theorm:

Theorom : $\displaystyle \int_0^\infty {x^m\over{x^n+1}}dx={{\pi \over n} \over sin(m+1({\pi \over n })) },n-m \ge2$

Proof:

using contour integral we can write : $$\oint_C {z^m \over {z^n+1}} \,dz$$

if $C$ choosed ppropriately The integrand has n first-order singularities,at the $n$ $n$-th roots of $-1$,these singular points are uniformly spaced around the unit circle in the complex plane. thus using euler formula we can write:

$$ -1=e^{i(1+2k)\pi}$$

singular points are located at:

$$ z_k=(-1)^{1 \over n}=e^{i({{1+2k} \over n})\pi},k=0,1,2,3,...,n-1 $$

for other values of k these same n points simply repeat.

now focus on one of these singular points , $k=0$.

pick C to enclose just that one singularity, at $z=z_0=e^{i{\pi \over n}}$

enter image description here

so the central angle of the wedge is $2\pi \over n$ and the singularity is at half that $\pi \over n$.

contour’s three portions are: $$ C_1=z=x,dz=dx,0 \le x \le T $$

$$ C_2=z=Te^{i\theta},dz=iTe^{i\theta}d\theta,0 \le x \le {2\pi \over n} $$

$$ C_3=z=re^{{i2\pi}\over n},dz=e^{{i2\pi}\over n}dr,0 \le r \le T $$

so:

$$ \oint_C {z^m \over {z^n+1}} \,dz= \int_0^T {x^m \over {x^n+1}}dx + \int_0^{2\pi \over n} {(Te^{i\theta})^m \over {(Te^{i\theta})^n+1}}iTe^{i\theta}d\theta+ \int_T^0 {(re^{{i2\pi}\over n})^m \over {(re^{{i2\pi}\over n})^n+1}}e^{{i2\pi}\over n}dr$$

$$=\int_0^T {x^m \over {x^n+1}}dx- \int_0^T { {r^me^{i(m+1){2\pi \over n}}} \over {r^n+1} }dr + \int_0^{2\pi \over n}{ {T^{m+1}e^{im\theta}} \over {T^{n}e^{in\theta}+1} }ie^{i\theta}d\theta $$

now clearly as $T \to \infty$ the $\theta$-integral goes to zero because $m+1<n$ also

$$\int_0^T { {r^me^{i(m+1){2\pi \over n}}} \over {r^n+1} }dr=e^{i(m+1){2\pi \over n}} \int_0^T {x^m \over {x^n+1}}dx $$

so as $T \to \infty$:

$$\oint_C {z^m \over {z^n+1}} \,dz=\int_0^\infty {x^m \over {x^n+1}}dx(1-e^{i(m+1){2\pi \over n}})$$

or as :

$$(1-e^{i(m+1){2\pi \over n}})=-2isin((m-1){\pi \over n})e^{i(m+1){2\pi \over n}}$$ we have:

$$\oint_C {z^m \over {z^n+1}} \,dz=-2isin((m=1){\pi \over n})e^{i(m+1){2\pi \over n}}\int_0^\infty {x^m \over {x^n+1}}dx $$

since we can write the integrand of the contour integral as a partial fraction expansion:

$$ {z^m \over {z^n+1}}={N_0 \over {z-z_0}}+{N_1 \over {z-z_1}}+...+{N_{n-1} \over {z-z_{n-1}}} $$

where the $N$'s are constants.integrating this expansion term-by-term:

$$ \oint_C {z^m \over {z^n+1}} \,dz=N_0\oint_C {dz \over {z-z_0}} $$

using cauchy’s first integral theorem all the other integrals are zero ,so the only singularity $C$ is $z_0$ ,now using cauchy’s second integral theorem with $f(z)=1$ we get:

$$ -2isin((m=1){\pi \over n})e^{i(m+1){2\pi \over n}}\int_0^\infty {x^m \over {x^n+1}}dx=2\pi iN_0 $$

and final step is calculate $N_0$:

$$ {(z-z_0)z^m \over {z^n+1}}=N_0+{N_1(z-z_0) \over {z-z_1}}+... $$

and if $z \to z_0$ then: $$ N_0=lim_{z \to z_0} {(z-z_0)z^m \over {z^n+1}}={0 \over 0} $$ thus using L’Hoˆpital’s rule: $$ N_0=lim_{z \to z_0} {(z-z_0)z^m \over {z^n+1}}={z_0^{m-n+1} \over n} $$

using $z_0=e^{i\pi \over n}$: $$ N_0=-{e^{i( {m+1 \over n} )\pi} \over n } $$ inserting this to result we finally get : $$ \bbox[5px,border:2px solid red] { \displaystyle \int_0^\infty {x^m\over{x^n+1}}dx={ {2\pi i}(-{e^{i( {m+1 \over n} )\pi} \over n }) \over {-2isin((m-1){\pi \over n})e^{i(m+1){2\pi \over n}}} } } $$ now in order to calculate what you asked: define $t=x^n$ then : $${dt \over dx} = nx^{n-1} $$ so : $$\int_0^\infty { t^{{m} \over n} \over t+1} ({dt \over nt^{{n-1}\over n} })=\int_0^\infty { t^{{{m+1} \over n}-1} \over {t+1}}dt $$

now define $$a={{m+1} \over n} $$ and finally get($0<a<1$): $$ \bbox[5px,border:2px solid green] { \displaystyle \int_0^\infty {x^{a-1}\over{x+1}}dx= {\pi \over sin(a\pi)} } $$

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