# Calculation of integral using Gamma function when the imaginary part is zero

Consider the following expression of Gamma function

$$\frac{\Gamma(z)}{p^z}=\int_{0}^{\infty}e^{-pt}t^{z-1}dt \ \ \ \ \ \ \ \ \ (1)$$ where $Re(z)>0$ and $Re(p)>0$.

In Lebedevs book "special functions and their applications" he uses that integral to calculate

$$\int_{0}^{\infty}e^{2izs}s^{\alpha-1}ds=\frac{\Gamma(\alpha)}{(-2iz)^{\alpha}} \ \ \ \ \ \ \ \ (2)$$

but what about case when $z=x$ is real? This means $Re(p)=0$ in $(1)$ expression.

P.S. The Formula $(2)$ must be valid for Real $z$ because of final result of calculation (asymptotic expression of Hankels' function $H_{\nu}^{(1)}(z)$).

I just want to know how to integrate (2) integral if z=x is real variable.

• Hint: integration by parts and analytic continuation. – Fan Zheng Nov 14 '15 at 20:41

The identity can be deduced by the Euler integral by homotopy invariance of path integrals.

The function $e^{-z}z^{\alpha}$ is holomorphic on $\mathbb{C}\setminus \{\operatorname{Re}z\le0,\operatorname{Im}z=0 \}$, so the value of the path integral along the boundary of the domain $\{z: \operatorname{Re}z>0,\operatorname{Im}z>0, \epsilon< |z|< \rho \}$ is zero. Hence we get $$\int_\epsilon^{\rho}i^\alpha e^{-it}t^{\alpha-1}dt=\int_\epsilon^{\rho}e^{-t}t^{\alpha-1}dt+ \int_0^{\pi/2}ie^{-\rho e^{it} }(\rho e^{it})^{\alpha}dt- \int_0^{\pi/2}ie^{-\epsilon e^{it}}(\epsilon e^{it} )^{\alpha}dt$$

We consider separately the three integrals on the RHS.

• The first integral, of course, converges to the Euler integral for $\Gamma(\alpha)$ as $\epsilon\to0$ and $\rho\to+ \infty$, provided $\operatorname{Re}\alpha>0$.

• The second integrand has absolute value $e^{-\rho\cos(t) -\operatorname{Im}\alpha t}\; \rho^{ \operatorname{Re}\alpha}$, and it is a bit singular at $t=\pi/2$; to evaluate the corresponding integral, it is convenient to spit it further in the integrals over $[0,x]$ and $[x,\pi/2]$ with a free $x$, optimizing then the bound over the choice of $x$. This way one gets a bound for this integral of order $O\Big(\rho ^{\operatorname{Re}\alpha -1} \log(\rho) \Big)$, which is $o(1)$ as $\rho\to+ \infty$, provided $\operatorname{Re}\alpha <1$.

• The third integrand has absolute value $e^{-\cos(t) \epsilon-\operatorname{Im}\alpha t}\; \epsilon^{ \operatorname{Re}\alpha}=O(\epsilon^{ \operatorname{Re}\alpha})$.

We conclude that for $0< \operatorname{Re}\alpha<1$ the integral on the LHS converges, and

$$\int_0^{+\infty} e^{-it}t^{\alpha-1}dt=\frac{\Gamma(\alpha)}{i^\alpha},$$

which is the wanted identity for $z=-1/2$. For any real $z<0$, with a linear change of variable, $t=(-2z)s$ we plainly get

$$\int_0^{+\infty} e^{2izs}s^{\alpha-1}dt=\frac{\Gamma(\alpha)}{(-2iz)^\alpha}.$$

Finally, for a real positive $z>0$, we take complex conjugate to both sides to the latter identity, written for $-z$ and $\overline \alpha$, which yields to the identity for $z$ and $\alpha$.

let $z$ be a real number, $0<\alpha<1$, $L>0$ and start from the proper integral

$$\int_{0}^{L}e^{2izs}s^{\alpha-1}ds=\frac{\Gamma(\alpha)-\Gamma(\alpha,-2iLz)}{(-2iz)^{\alpha}}$$

use the asymptotics for the incomplete Gamma function: $$\quad\lim_{L\rightarrow\infty}\frac{\Gamma(\alpha,-2iLz)}{(-2iLz)^{\alpha-1}e^{2iLz}}=1\Rightarrow \lim_{L\rightarrow\infty}\Gamma(\alpha,-2iLz)=0$$

$$\Rightarrow\lim_{L\rightarrow\infty}\int_{0}^{L}e^{2izs}s^{\alpha-1}ds=\frac{\Gamma(\alpha)}{(-2iz)^{\alpha}}$$