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Consider $\oint e^{-|a|z}z^{s-1}$, taken around the contour consisting of the line from $\epsilon$ to $R$, the circular arc of radius $R$ where $arg(z)$ goes from $0$ to $arg(a)$, the line defined by $te^{iarg(a)}$, where $t$ goes from $R$ to $\epsilon$, and the circular arc of radius $\epsilon$ required to close the contour.

I want to use this integral to show that $\int_{0}^{\infty}e^{-at}t^{s-1}dt=a^{-s}\Gamma(s)$, where $a$ is in general complex and we have $Re(a)>0$ and $Re(s)>0$.

Now, the segment on the real axis tends to the integral equired as $\epsilon$ vanishes and $R$ tends to infinity.

The segment on the arc of radius $\epsilon$ is bounded by $K|arg(a)|\epsilon ^{Re(s)}$, where $K$ is constant (by application of the ML bound), and so vanishes as $\epsilon$ vanishes.

The integral along the straight segment can be easily evaluated as $-|a|^{s-1}\Gamma(s)$

Also, as this contour encloses no singularities, the total integral is $0$ by Cauchy's Theorem.

My issue is that I can't seem to show that the integral along the arc of radius $R$ vanishes. Clearly, to give the correctresult this must be the case, but I just can't seem to show it. It's been suggested to me to use an analog of Jordan's Lemma, but still I can't seem to figure it out. I can't seem to express the integrand in the form $e^{i\alpha z}f(z)$, with $\alpha>0$ as required by the Lemma. I'm also a bit worried about the fact that this isn't a full semicircular arc, and we don't know whether $a$ lies in the upper or lower half plane.

Any help would be much appreciated, thanks!

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We need not use contour integration to show that

$$\int_0^\infty e^{-|a|t}t^{s-1}\,dt=|a|^{-s}\Gamma(s) \tag 1$$

Rather, we note that the integration path is on the real line. Hence, enforcing the substitution $t\to t/|a|$ yields

$$\begin{align} \int_0^\infty e^{-|a|t}t^{s-1}\,dt&=\int_0^\infty e^{-t}\left(\frac{t}{|a|}\right)^{s-1}\,\frac{1}{|a|}\,dt\\\\ &=|a|^{-s}\int_0^\infty e^{-t}t^{s-1}\,dt\\\\ &=|a|^{-s} \Gamma(s) \end{align}$$

as was to be shown.

It remains to be shown that $\int_0^\infty e^{-at}\,t^{s-1}\,dt=a^{-s}\Gamma(s)$ for $|\arg(a)|<\pi/2$. It is to that end that we proceed.


Contour Integral

Let $f(z)=e^{-|a|z}z^{s-1}$. Note that $f(z)$ has a branch point at $z=0$. We choose the branch cut from $z=0$ to $z=-\infty$ along the negative real axis.

Then, $f(z)$ is analytic on this Riemann sheet and Cauchy's Integral Theorem guarantees that for any rectifiable closed curve, $C$, that does not intersect the chosen branch cut

$$\oint_C f(z)\,dz=0$$

We choose $C$ to be the contour as described in the OP. Therefore we write

$$\begin{align} 0&=\oint_C f(z)\,dz\\\\ &=\int_\epsilon^R e^{-|a|x}x^{s-1}\,dx \tag 2\\\\ &+\int_0^{\arg(a)} e^{-|a|Re^{i\phi}}\,(Re^{i\phi})^{s-1}\,iRe^{i\phi}\,d\phi\tag 3\\\\ &+\int_R^\epsilon e^{-|a|te^{i\arg(a)}}\,(te^{i\arg(a)})^{s-1}\,e^{i\arg(a)}\,dt\tag 4\\\\ &+\int_{\arg(a)}^0 e^{-|a|\epsilon e^{i\phi}}\,(\epsilon e^{i\phi})^{s-1}\,i\epsilon e^{i\phi}\,d\phi\tag 5 \end{align}$$

We assume that $|\arg(a)|< \pi/2$, else the integral in $(4)$ does not converge as $R\to \infty$.


Under this assumption, we see that the integral in $(2)$ becomes

$$\lim_{(\epsilon,R)\to (0,\infty)}\int_\epsilon^R e^{-|a|x}x^{s-1}\,dx=\int_0^\infty e^{-|a|x}x^{s-1}\,dx$$


For the integral in $(3)$ we have

$$\begin{align} \left|\int_0^{\arg(a)} e^{-|a|Re^{i\phi}}(Re^{i\phi})^{s-1}\,iRe^{i\phi}\,d\phi\right|&\le R^{\text{Re}(s)}\int_0^{|\arg(a)|} e^{-|a|R\cos(\phi)}\,d\phi\\\\ &\le R^{\text{Re}(s)}\int_0^{|\arg(a)|} e^{-|a|R(1-2\phi /\pi)}\,d\phi\\\\ &=\frac{\pi}{2|a|}R^{\text{Re}(s)-1}\left(e^{-|a|R\left(1-\frac 2\pi|\arg(a)|\right)} -e^{-|a|R}\right)\\\\ &\to 0\,\,\text{as}\,\,R\to \infty \end{align}$$


The integral in $(4)$ becomes

$$\lim_{(\epsilon,R)\to (0,\infty)}\int_R^\epsilon e^{-|a|te^{i\arg(a)}}\,(te^{i\arg(a)})^{s-1}\,e^{i\arg(a)}\,dt=-e^{is\arg(a)}\int_0^\infty e^{-at}\,t^{s-1}\,dt$$


And the integral in $(5)$ vanishes as $\epsilon\to 0$.


Putting everything together reveals

$$\int_0^\infty e^{-at}\,t^{s-1}\,dt=\left(e^{i\arg(a)}\right)^{-s}\int_0^\infty e^{-|a|x}x^{s-1}\,dx=a^{-s}\Gamma(s)$$

which was to be shown!

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