Homework 05

1. Integral, differentiability and decay rate

Compute this generalization of an integral we did in class:

\[ I_{n,a}(k) = \int^{∞}_{−∞} \frac{ e^{ikx} }{ (x^2 + a^2)^n} dx \]

where $k, a \in {\mathbb{R}}, a > 0$ and $n \in {\mathbb{Z}}, n \geq 1$. What is the degree of differentiability of $I_{n,a}(k)$ with respect to $k$? How does this relate to the decay of the integrand?

Compute the Integral:

Complex function $f(z) = \frac{e^{ikz}}{(z^2 + a^2)^n}$ has poles at $z = \pm i a$.

The residue at $z = + i a$ pole

\[ \text{Res}(f, ia) = \frac{1}{(n-1)!} \lim_{z \to ia} \frac{d^{n-1}}{dz^{n-1}} \left[ (z - ia)^n f(z) \right] \]

Construct a semicircular contour in the upper half-plane that consists of a line segment $C_1$ from $−R$ to $R$ and a semicircular arc $C_2$ of radius $R$ centered at the origin.

The integral over the contour $C$ is:

\[ \int_C f(z) dz = 2\pi i \times \text{Res}(f, ia) \]

As $R \to \infty$, the integral over $C_2$ vanishes for $k>0$ due to the exponential decay of $e^{ikz}$ in the upper half-plane.

The original integral $I_{n,a}(k)$ is equal to the integral over $C_1$:

\[ I_{n,a}(k) = 2\pi i \times \text{Res}(f, ia) \]

Degree of Differentiability

\[ I_{n,a}'(k) = \int^{∞}_{−∞} \frac{ ix e^{ikx} }{ (x^2 + a^2)^n} dx \]

The corresponding complex function

\[ f_k(z) = \frac{i z e^{ikz}}{(z^2 + a^2)^n} \]

has poles at $z = \pm i a$.

The function $I_{n,a}(k)$ will be infinitely differentiable with respect to $k$ because the exponential function $e^{ikx}$ is smooth and the denominator does not depend on $k$.

Relation to the Decay of the Integrand:

The decay of the integrand as $x \to \pm\infty$ is crucial for the convergence of the integral. The factor $(x^2 + a^2)^{-n}$ ensures that the integrand decays sufficiently fast at infinity for the integral to converge.

The faster decay of the integrand at infinity, which is ensured by a higher $n$, means that $I_{n,a}(k)$ will be more regular (i.e., higher degree of differentiability).

2. Integral of a keyhole contour

Compute the following integral by a choice of keyhole contour, where $a, b, c ∈ {\mathbb{R}}$ and $b, c > 0$:

\[ I(a, b, c) = \int^∞_{−∞} dx \frac{\log(x^2 + c^2)}{ (x − a)^2 + b^2} \]

The complex function

\[ f(z) = \frac{\log(z^2 + c^2)}{ (z − a)^2 + b^2} = \frac{\log(z+ ic)}{ (z − a)^2 + b^2} + \frac{\log(z - ic)}{ (z − a)^2 + b^2} \]

may have branch point $𝑧 = \pm 𝑖c$ and have poles at $a \pm ib$.

Consider the contour integral $\oint_C dz f(z)$in the complex plane where $𝐶$ is a semicircle of radius $R$ in the upper half-plane with a detour down and up the imaginary axis about the branch point $z=ic$.

The function $\log(z + ic)$ is holomorphic in the upper half plane if we choose its branch cut to lie in the lower half plane, so the integrand is holomorphic inside the contour of $C$ except for a simple pole at $a + ib$.

\[ \oint_C dz \frac{\log(z+ ic)}{ (z − a)^2 + b^2} = 2 \pi i \frac{\log(a+ib+ic)}{2ib} = \frac{\pi}{b} \log(a+ib+ic) \]

The function $\log(z - ic)$ has a branch cut in the upper half plane. We have

\[ \oint_C dz \frac{\log(z - ic)}{ (z − a)^2 + b^2} = \frac{\pi}{b} \log(a + ib - ic) + \int_{ic}^{\infty} \frac{ \log (iy-ic+\epsilon) - \log (iy-ic+\epsilon)}{ (z-a)^2+b^2} \]

Since the contribution from the big semi-circle tends to zero as $R → ∞$ and from the small circle around $ic$ as its radius tends to zero. And

\[ \int_{ic}^{\infty} \frac{ \log (iy-ic+\epsilon) - \log (iy-ic+\epsilon)}{ (z-a)^2+b^2} = -2 \pi \int_c^{\infty}dy \frac{1}{(iy-a)^2+b^2} = - \frac{\pi}{b}\log \frac{a+ib-ic}{a-ib-ic} \]

\[ \begin{align*} I(a, b, c) &= \int^∞_{−∞} dx \frac{\log(x^2 + c^2)}{ (x − a)^2 + b^2} \\ &= \frac{\pi}{b} \log(a+ib+ic) + \frac{\pi}{b} \log(a+ib-ic) - - \frac{\pi}{b}\log \frac{a+ib-ic}{a-ib-ic} \\ &= \frac{\pi}{b} \log(a^2+(b+c)^2) \end{align*} \]

3. Integral and Laurent expansion

Consider the integral

\[ \int_C dz \frac{z + 2}{ z^2 − 9} \]

where $C$ is a positively-oriented circle of radius 4. Compute this integral by taking the sum of the residues inside the circle, and then again by computing the “residue at $∞$” meaning using the outer Laurent expansion

The integrand

\[ f(z) = \frac{z + 2}{z^2 - 9} \]

have simple poles at $z =\pm 3$.

\[ \text{Res}(f, 3) = \lim_{z \to 3} (z - 3) \frac{z + 2}{z^2 - 9} = \frac{5}{6} \]

\[ \text{Res}(f, -3) = \lim_{z \to -3} (z + 3) \frac{z + 2}{z^2 - 9} = \frac{1}{6} \]

By the residue theorem, the integral over the contour $C$ is:

\[ \int_C f(z) , dz = 2\pi i \times (\text{Res}(f, 3) + \text{Res}(f, -3)) = 2\pi i \]

Using the outer Laurent expansion

\[ \frac{z + 2}{z^2 - 9} = \frac{1}{z} + O[\frac{1}{z^2}] \]

So using residue at $∞$

\[ \text{Res}(f, ∞) = 1 \] \[ \int_C f(z) , dz = 2\pi i \times \text{Res}(f, ∞) = 2\pi i \]

4. Analytically continuing

Compute the integral

\[ \int^{2π}_{0} \frac{dθ}{2 − \cos θ } \]

by analytically continuing the integrand and then using residues.

Perform a change of variables to convert the integral into a contour integral in the complex plane.

Let $z = e^{i\theta}$. Then, $dz = ie^{i\theta}d\theta = izd\theta$ and $d\theta = \frac{dz}{iz}$. Also, $\cos \theta = \frac{1}{2}(e^{i\theta} + e^{-i\theta}) = \frac{1}{2}(z + \frac{1}{z})$.

Substituting these into the integral, we get:

\[ \int^{2π}_{0} \frac{dθ}{2 − \cos θ } = \int_{|z|=1} \frac{1}{2 - \frac{1}{2}(z + \frac{1}{z})} \times \frac{dz}{iz} \]

Simplifying this, we have:

\[ \int_{|z|=1} \frac{dz}{2iz - \frac{1}{2}i(z^2 + 1)} \]

To find the poles of the integrand, we set the denominator to zero:

\[ iz - \frac{1}{2}i(z^2 + 1) =0 \]

\[ z_1=2-\sqrt{3}, z_2=2+\sqrt{3} \]

$z_1$ is within the contour, so

\[ \int_{|z|=1} \frac{1}{2 - \frac{1}{2}(z + \frac{1}{z})} \times \frac{dz}{iz} = 2 \pi i \frac{2}{i(z_2-z_1)} = \frac {2 \pi}{\sqrt 3} \]

5. Gamma function

The Gamma function is usually defined as

\[ Γ(z) = \int^∞_0 t^{z−1}e^{−t}dt \]

(This kind of integral is known as a Mellin transform, in this case of $e^{−t}$, and we can equivalently write it as a two-sided Laplace transform of $e^{−e^s}$ by taking $s = \log t$.) Verify this integral exists for $\Re(z) > 0$ and that

\[ zΓ(z) = Γ(z + 1) \] \[ Γ(n + 1) = n! \]

The first relation can be iterated to analytically continue the $Γ$ function to the whole complex plane, except for poles at the nonpositive integers. However, there is a way to analytically continue it all at once:

Split the integral into two parts $\int^∞_ 1 +\int^1_0$ . Show that the first part yields an entire function in $z$, and that the second part equals

\[ \sum_{n=0}^\infty \frac{(−1)^n}{n!} \frac{1}{ z + n} \]

which is an entire meromorphic function with poles at the non-positive integers, thus giving an analytic continuation of $Γ$ as a meromorphic function on all of ${\mathbb{C}}$.

To verify the properties of the Gamma function and its analytic continuation, we’ll break down the problem into parts.

Existence

We prove this integral converges absolutely.

\[ \int\limits_0^\infty |t^{z-1}e^{-t}| dt = \int\limits_0^\infty t^{\operatorname{Re} z-1}e^{-t} dt \]

For $z \in {\mathbb{R}}$,

\[ \int\limits_0^\infty t^{z-1}e^{-t}dt = \int\limits_0^1 t^{z-1}e^{-t}dt + \int\limits_1^\infty t^{z-1}e^{-t}dt \]

we have for large $N \le t,$ $t^{z-1}e^{-t}\le e^{-t/2}$

\[ 0\le t\le N \implies t^{z-1}e^{-t}\le t^{z-1} \;,\;\;\text{and}\;\; \int\limits_0^N t^{z-1}dt=\left.\frac{t^z}z\right|_0^N=\frac{N^z}{z} \]

\[ N \le t\implies t^{z-1}e^{-t}\le e^{-t/2} \;,\;\;\text{and}\;\; \int\limits_N^\infty e^{-t/2} dt=\left. -2 e^{-t/2}\right|_N^\infty= 2e^{-N/2} \]

\[ \int\limits_0^\infty |t^{z-1}e^{-t}| dt \le \infty \]

Therefore, this integral exists for $\Re(z) > 0$.

$zΓ(z) = Γ(z + 1)$

By definition:

\[ z\Gamma(z) = z \int_{0}^{\infty} t^{z-1}e^{-t} dt \]

Using integration by parts, we get:

\[ \begin{aligned} z\Gamma(z) &= \left. t^z e^{-t} \right| _{0}^{\infty} - \int_{0}^{\infty} t^{z} (-e^{-t}) dt \\ &= \int_{0}^{\infty} t^z e^{-t} dt \\ &= \Gamma(z + 1) \end{aligned} \]

$Γ(n + 1) = n!$

\[ {\displaystyle {\begin{aligned}\Gamma (1)&=\int _{0}^{\infty }t^{1-1}e^{-t}\,dt\\&=\int _{0}^{\infty }e^{-t}\,dt\\&=1.\end{aligned}}} \]

By induction we have $Γ(n + 1) = n!$

Splitting

Now, to analytically continue the Gamma function, we split the integral into two parts:

\[ \Gamma(z) = \int_{1}^{\infty} t^{z-1}e^{-t} dt + \int_{0}^{1} t^{z-1}e^{-t} dt \]

The first part $\int_{1}^{\infty} t^{z-1}e^{-t} dt = (\int_{1}^{N} + \int_{N}^{\infty}) t^{z-1}e^{-t} dt$ converges for all $z$ and is an entire function in $z$.
The second part $\int_{0}^{1} t^{z-1}e^{-t} dt$ ) can be expressed as a power series using the Taylor expansion of $e^{-t}$ :

\[ \int_{0}^{1} t^{z-1}e^{-t}dt = \int_{0}^{1} t^{z-1} \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} t^n dt = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \int_{0}^{1} t^{z + n - 1} dt = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \frac{1}{z + n} \]

This series converges for ( z ) not a non-positive integer and provides an analytic continuation of $\Gamma$ as a meromorphic function on all of $\mathbb{C}$, with poles at the non-positive integers.

6. Gamma relation

Prove the relation

\[ Γ(z)Γ(1 − z) = \frac{π} {\sin πz} \]

by combining the two Gamma integrals into

\[ Γ(z)Γ(1 − z) = \int^∞_0 dt \frac{t^{z−1}}{t + 1} \]

and evaluating this by residues.

\[ \Gamma(z) = \int_{0}^{\infty} t^{z-1}e^{-t} dt \]

For fixed $t$

\[ \Gamma(1 - z) = \int_{0}^{\infty} t^{-z}e^{-t} dt = \int_{0}^{\infty} u^{-z}e^{-u} du = t \int_{0}^{\infty} (vt)^{-z}e^{-vt} dv \]

By combining these two

\[ Γ(z)Γ(1 − z) = \int^∞_0\int^∞_0 e^{-t(1+v)} v^{-z} dvdt = \int^∞_0 dv \frac{v^{z−1}}{v + 1} = \int^∞_0 dt \frac{t^{z−1}}{t + 1} = \int^∞_{-∞} dt \frac{e^{zx}}{ 1+e^x } \]

Complex function

\[ \frac{e^{zx}}{ 1+e^x } \]

have pole at $\pi i$. Consider a contour $C_R$ in the complex plane with vertices at $R$, $−R$, $R+2πi$, and $−R+2πi$, as $R$ tends to infinity. We have

\[ {\displaystyle \int _{C_{R}}{\frac {e^{xz}}{1+e^{x}}} dx=-2\pi ie^{z \pi i}} \] \[ \int _{C_{R3}}{f(x,z)} dx = -e^{2\pi i z} \int _{C_{R1}}{f(x,z)} dx \]

The right and left vertical sides of the rectangle tend to 0 as $R \to \infty$ for $z \in (0,1)$.

\[ \int^∞_{-∞} dt \frac{e^{zx}}{ 1+e^x } = \frac{-2\pi ie^{z \pi i}}{1-e^{2\pi i z}} = \frac{\pi}{\sin \pi z} \]

By analytic continuation, this relation is true for all $z \in {\mathbb{C}}/\ {\mathbb{Z}}$

--- title: Homework 05 format: html: default typst: output-file: phys231_hw05 --- \newcommand\R{{\mathbb{R}}} \newcommand\Z{{\mathbb{Z}}} \newcommand\C{{\mathbb{C}}} ## 1. Integral, differentiability and decay rate > Compute this generalization of an integral we did in class: > > $$ > I_{n,a}(k) = \int^{∞}_{−∞} \frac{ e^{ikx} }{ (x^2 + a^2)^n} dx > $$ > > where $k, a \in \R, a > 0$ and $n \in \Z, n \geq 1$. What is the degree of differentiability of $I_{n,a}(k)$ with respect to $k$? How does this relate to the decay of the integrand? ### Compute the Integral: Complex function $f(z) = \frac{e^{ikz}}{(z^2 + a^2)^n}$ has poles at $z = \pm i a$. The residue at $z = + i a$ pole $$ \text{Res}(f, ia) = \frac{1}{(n-1)!} \lim_{z \to ia} \frac{d^{n-1}}{dz^{n-1}} \left[ (z - ia)^n f(z) \right] $$ Construct a semicircular contour in the upper half-plane that consists of a line segment $C_1$ from $−R$ to $R$ and a semicircular arc $C_2$ of radius $R$ centered at the origin. The integral over the contour $C$ is: $$ \int_C f(z) dz = 2\pi i \times \text{Res}(f, ia) $$ As $R \to \infty$, the integral over $C_2$ vanishes for $k>0$ due to the exponential decay of $e^{ikz}$ in the upper half-plane. The original integral $I_{n,a}(k)$ is equal to the integral over $C_1$: $$ I_{n,a}(k) = 2\pi i \times \text{Res}(f, ia) $$ ### Degree of Differentiability $$ I_{n,a}'(k) = \int^{∞}_{−∞} \frac{ ix e^{ikx} }{ (x^2 + a^2)^n} dx $$ The corresponding complex function $$ f_k(z) = \frac{i z e^{ikz}}{(z^2 + a^2)^n} $$ has poles at $z = \pm i a$. The function $I_{n,a}(k)$ will be infinitely differentiable with respect to $k$ because the exponential function $e^{ikx}$ is smooth and the denominator does not depend on $k$. ### Relation to the Decay of the Integrand: The decay of the integrand as $x \to \pm\infty$ is crucial for the convergence of the integral. The factor $(x^2 + a^2)^{-n}$ ensures that the integrand decays sufficiently fast at infinity for the integral to converge. The faster decay of the integrand at infinity, which is ensured by a higher $n$, means that $I_{n,a}(k)$ will be more regular (i.e., higher degree of differentiability). ## 2. Integral of a keyhole contour > Compute the following integral by a choice of keyhole contour, where $a, b, c ∈ \R$ and $b, c > 0$: > > $$ > I(a, b, c) = \int^∞_{−∞} dx \frac{\log(x^2 + c^2)}{ (x − a)^2 + b^2} > $$ The complex function $$ f(z) = \frac{\log(z^2 + c^2)}{ (z − a)^2 + b^2} = \frac{\log(z+ ic)}{ (z − a)^2 + b^2} + \frac{\log(z - ic)}{ (z − a)^2 + b^2} $$ may have branch point $𝑧 = \pm 𝑖c$ and have poles at $a \pm ib$. Consider the contour integral $\oint_C dz f(z)$in the complex plane where $𝐶$ is a semicircle of radius $R$ in the upper half-plane with a detour down and up the imaginary axis about the branch point $z=ic$. ![](images/keyhole_contour.png) The function $\log(z + ic)$ is holomorphic in the upper half plane if we choose its branch cut to lie in the lower half plane, so the integrand is holomorphic inside the contour of $C$ except for a simple pole at $a + ib$. $$ \oint_C dz \frac{\log(z+ ic)}{ (z − a)^2 + b^2} = 2 \pi i \frac{\log(a+ib+ic)}{2ib} = \frac{\pi}{b} \log(a+ib+ic) $$ The function $\log(z - ic)$ has a branch cut in the upper half plane. We have $$ \oint_C dz \frac{\log(z - ic)}{ (z − a)^2 + b^2} = \frac{\pi}{b} \log(a + ib - ic) + \int_{ic}^{\infty} \frac{ \log (iy-ic+\epsilon) - \log (iy-ic+\epsilon)}{ (z-a)^2+b^2} $$ Since the contribution from the big semi-circle tends to zero as $R → ∞$ and from the small circle around $ic$ as its radius tends to zero. And $$ \int_{ic}^{\infty} \frac{ \log (iy-ic+\epsilon) - \log (iy-ic+\epsilon)}{ (z-a)^2+b^2} = -2 \pi \int_c^{\infty}dy \frac{1}{(iy-a)^2+b^2} = - \frac{\pi}{b}\log \frac{a+ib-ic}{a-ib-ic} $$ So $$ \begin{align*} I(a, b, c) &= \int^∞_{−∞} dx \frac{\log(x^2 + c^2)}{ (x − a)^2 + b^2} \\ &= \frac{\pi}{b} \log(a+ib+ic) + \frac{\pi}{b} \log(a+ib-ic) - - \frac{\pi}{b}\log \frac{a+ib-ic}{a-ib-ic} \\ &= \frac{\pi}{b} \log(a^2+(b+c)^2) \end{align*} $$ ## 3. Integral and Laurent expansion > Consider the integral > > $$ > \int_C dz \frac{z + 2}{ z^2 − 9} > $$ > > where $C$ is a positively-oriented circle of radius 4. Compute this integral by taking the sum of the residues inside the circle, and then again by computing the “residue at $∞$” meaning using the outer Laurent expansion The integrand $$ f(z) = \frac{z + 2}{z^2 - 9} $$ have simple poles at $z =\pm 3$. $$ \text{Res}(f, 3) = \lim_{z \to 3} (z - 3) \frac{z + 2}{z^2 - 9} = \frac{5}{6} $$ $$ \text{Res}(f, -3) = \lim_{z \to -3} (z + 3) \frac{z + 2}{z^2 - 9} = \frac{1}{6} $$ By the residue theorem, the integral over the contour $C$ is: $$ \int_C f(z) , dz = 2\pi i \times (\text{Res}(f, 3) + \text{Res}(f, -3)) = 2\pi i $$ Using the outer Laurent expansion $$ \frac{z + 2}{z^2 - 9} = \frac{1}{z} + O[\frac{1}{z^2}] $$ So using residue at $∞$ $$ \text{Res}(f, ∞) = 1 $$ $$ \int_C f(z) , dz = 2\pi i \times \text{Res}(f, ∞) = 2\pi i $$ ## 4. Analytically continuing > Compute the integral > > $$ > \int^{2π}_{0} \frac{dθ}{2 − \cos θ } > $$ > > by analytically continuing the integrand and then using residues. Perform a change of variables to convert the integral into a contour integral in the complex plane. Let $z = e^{i\theta}$. Then, $dz = ie^{i\theta}d\theta = izd\theta$ and $d\theta = \frac{dz}{iz}$. Also, $\cos \theta = \frac{1}{2}(e^{i\theta} + e^{-i\theta}) = \frac{1}{2}(z + \frac{1}{z})$. Substituting these into the integral, we get: $$ \int^{2π}_{0} \frac{dθ}{2 − \cos θ } = \int_{|z|=1} \frac{1}{2 - \frac{1}{2}(z + \frac{1}{z})} \times \frac{dz}{iz} $$ Simplifying this, we have: $$ \int_{|z|=1} \frac{dz}{2iz - \frac{1}{2}i(z^2 + 1)} $$ To find the poles of the integrand, we set the denominator to zero: $$ iz - \frac{1}{2}i(z^2 + 1) =0 $$ $$ z_1=2-\sqrt{3}, z_2=2+\sqrt{3} $$ $z_1$ is within the contour, so $$ \int_{|z|=1} \frac{1}{2 - \frac{1}{2}(z + \frac{1}{z})} \times \frac{dz}{iz} = 2 \pi i \frac{2}{i(z_2-z_1)} = \frac {2 \pi}{\sqrt 3} $$ ## 5. Gamma function > The Gamma function is usually defined as > > $$ > Γ(z) = \int^∞_0 t^{z−1}e^{−t}dt > $$ > > (This kind of integral is known as a Mellin transform, in this case of $e^{−t}$, and we can equivalently write it as a two-sided Laplace transform of $e^{−e^s}$ by taking $s = \log t$.) Verify this integral exists for $\Re(z) > 0$ and that > > $$ > zΓ(z) = Γ(z + 1) > $$ > $$ > Γ(n + 1) = n! > $$ > > The first relation can be iterated to analytically continue the $Γ$ function to the whole complex plane, except for poles at the nonpositive integers. However, there is a way to analytically continue it all at once: > > Split the integral into two parts $\int^∞_ 1 +\int^1_0$ . Show that the first part yields an entire function in $z$, and that the second part equals > > $$ > \sum_{n=0}^\infty \frac{(−1)^n}{n!} \frac{1}{ z + n} > $$ > > which is an entire meromorphic function with poles at the non-positive integers, thus giving an analytic continuation of $Γ$ as a meromorphic function on all of $\C$. To verify the properties of the Gamma function and its analytic continuation, we'll break down the problem into parts. ### Existence We prove this integral converges absolutely. $$ \int\limits_0^\infty |t^{z-1}e^{-t}| dt = \int\limits_0^\infty t^{\operatorname{Re} z-1}e^{-t} dt $$ For $z \in \R$, $$ \int\limits_0^\infty t^{z-1}e^{-t}dt = \int\limits_0^1 t^{z-1}e^{-t}dt + \int\limits_1^\infty t^{z-1}e^{-t}dt $$ we have for large $N \le t,$ $t^{z-1}e^{-t}\le e^{-t/2}$ $$ 0\le t\le N \implies t^{z-1}e^{-t}\le t^{z-1} \;,\;\;\text{and}\;\; \int\limits_0^N t^{z-1}dt=\left.\frac{t^z}z\right|_0^N=\frac{N^z}{z} $$ $$ N \le t\implies t^{z-1}e^{-t}\le e^{-t/2} \;,\;\;\text{and}\;\; \int\limits_N^\infty e^{-t/2} dt=\left. -2 e^{-t/2}\right|_N^\infty= 2e^{-N/2} $$ So $$ \int\limits_0^\infty |t^{z-1}e^{-t}| dt \le \infty $$ Therefore, this integral exists for $\Re(z) > 0$. ### $zΓ(z) = Γ(z + 1)$ By definition: $$ z\Gamma(z) = z \int_{0}^{\infty} t^{z-1}e^{-t} dt $$ Using integration by parts, we get: $$ \begin{aligned} z\Gamma(z) &= \left. t^z e^{-t} \right| _{0}^{\infty} - \int_{0}^{\infty} t^{z} (-e^{-t}) dt \\ &= \int_{0}^{\infty} t^z e^{-t} dt \\ &= \Gamma(z + 1) \end{aligned} $$ ### $Γ(n + 1) = n!$ $$ {\displaystyle {\begin{aligned}\Gamma (1)&=\int _{0}^{\infty }t^{1-1}e^{-t}\,dt\\&=\int _{0}^{\infty }e^{-t}\,dt\\&=1.\end{aligned}}} $$ By induction we have $Γ(n + 1) = n!$ ### Splitting Now, to analytically continue the Gamma function, we split the integral into two parts: $$ \Gamma(z) = \int_{1}^{\infty} t^{z-1}e^{-t} dt + \int_{0}^{1} t^{z-1}e^{-t} dt $$ * The first part $\int_{1}^{\infty} t^{z-1}e^{-t} dt = (\int_{1}^{N} + \int_{N}^{\infty}) t^{z-1}e^{-t} dt$ converges for all $z$ and is an entire function in $z$. * The second part $\int_{0}^{1} t^{z-1}e^{-t} dt$ ) can be expressed as a power series using the Taylor expansion of $e^{-t}$ : $$ \int_{0}^{1} t^{z-1}e^{-t}dt = \int_{0}^{1} t^{z-1} \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} t^n dt = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \int_{0}^{1} t^{z + n - 1} dt = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \frac{1}{z + n} $$ This series converges for ( z ) not a non-positive integer and provides an analytic continuation of $\Gamma$ as a meromorphic function on all of $\mathbb{C}$, with poles at the non-positive integers. ## 6. Gamma relation > Prove the relation > > $$ > Γ(z)Γ(1 − z) = \frac{π} {\sin πz} > $$ > > by combining the two Gamma integrals into > > $$ > Γ(z)Γ(1 − z) = \int^∞_0 dt \frac{t^{z−1}}{t + 1} > $$ > > and evaluating this by residues. $$ \Gamma(z) = \int_{0}^{\infty} t^{z-1}e^{-t} dt $$ For fixed $t$ $$ \Gamma(1 - z) = \int_{0}^{\infty} t^{-z}e^{-t} dt = \int_{0}^{\infty} u^{-z}e^{-u} du = t \int_{0}^{\infty} (vt)^{-z}e^{-vt} dv $$ By combining these two $$ Γ(z)Γ(1 − z) = \int^∞_0\int^∞_0 e^{-t(1+v)} v^{-z} dvdt = \int^∞_0 dv \frac{v^{z−1}}{v + 1} = \int^∞_0 dt \frac{t^{z−1}}{t + 1} = \int^∞_{-∞} dt \frac{e^{zx}}{ 1+e^x } $$ Complex function $$ \frac{e^{zx}}{ 1+e^x } $$ have pole at $\pi i$. Consider a contour $C_R$ in the complex plane with vertices at $R$, $−R$, $R+2πi$, and $−R+2πi$, as $R$ tends to infinity. We have $$ {\displaystyle \int _{C_{R}}{\frac {e^{xz}}{1+e^{x}}} dx=-2\pi ie^{z \pi i}} $$ $$ \int _{C_{R3}}{f(x,z)} dx = -e^{2\pi i z} \int _{C_{R1}}{f(x,z)} dx $$ The right and left vertical sides of the rectangle tend to 0 as $R \to \infty$ for $z \in (0,1)$. $$ \int^∞_{-∞} dt \frac{e^{zx}}{ 1+e^x } = \frac{-2\pi ie^{z \pi i}}{1-e^{2\pi i z}} = \frac{\pi}{\sin \pi z} $$ By analytic continuation, this relation is true for all $z \in \C /\ \Z$