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Chapter 34: Problem 6

Zeigen Sie f眉r \(\operatorname{Re} z \geq 0, z \neq 0\) die Beziehung \(\Gamma(\bar{z})=\overline{\Gamma(z)}\). (Diese Beziehung gilt tats盲chlich sogar f眉r alle \(z \in D(\Gamma)\).) Beweisen Sie damit $$ |\Gamma(\mathrm{ix})|^{2}=\frac{\pi}{x \sinh (\pi x)} $$ f眉r \(x \in \mathbb{R}_{\neq 0}\).

Short Answer

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Question: Prove that for Re(z) 鈮� 0 and z 鈮� 0, the relationship between the complex conjugate of z, 饾懅虆 , and the gamma function holds: 饾洡(饾懅虆) = 饾洡虆(z). Then use this relationship to prove that for x 鈭� 鈩� \ {0}, the following holds: |饾洡(ix)|虏 = 蟺/(x sinh(蟺x)).

Step by step solution

01

Prove the relationship between gamma function and complex conjugate

To show \(\Gamma(\bar{z}) = \overline{\Gamma(z)}\), we will use the definition of the gamma function: $$ \Gamma(z) = \int_0^\infty t^{z-1} e^{-t} dt, $$ where \(z\) is a complex number. Now, let's take the complex conjugate of the gamma function: $$ \overline{\Gamma(z)} = \overline{\int_0^\infty t^{z-1} e^{-t} dt}. $$ According to the properties of complex conjugates, we can move the conjugate inside of the integral: $$ \overline{\Gamma(z)} = \int_0^\infty \overline{t^{z-1} e^{-t}} dt. $$ Now, let's take the conjugate of \(t^{z-1}\) and \(e^{-t}\) individually: $$ \overline{t^{z-1}} = t^{\bar{z}-1} \quad \text{and} \quad \overline{e^{-t}} = e^{-t}. $$ Plug these back into the integral: $$ \overline{\Gamma(z)} = \int_0^\infty t^{\bar{z}-1} e^{-t} dt = \Gamma(\bar{z}). $$ So, we have proven that \(\Gamma(\bar{z}) = \overline{\Gamma(z)}\).
02

Prove the second statement using the relationship

Now that we've proven the relationship between the gamma function and complex conjugate, let's use it to prove the second statement: $$ |\Gamma(\mathrm{ix})|^2 = \frac{\pi}{x \sinh (\pi x)}, $$ for \(x \in \mathbb{R}_{\neq 0}\). First, let's evaluate the gamma function at \(\mathrm{ix}\) and its conjugate, \(\mathrm{-ix}\): $$ \Gamma(\mathrm{ix}) = \int_0^\infty t^{\mathrm{ix}-1} e^{-t} dt, $$ and $$ \Gamma(\mathrm{-ix}) = \int_0^\infty t^{\mathrm{-ix}-1} e^{-t} dt. $$ Since we know that \(\Gamma(\bar{z}) = \overline{\Gamma(z)}\), we also know that $$ \Gamma(\mathrm{-ix}) = \overline{\Gamma(\mathrm{ix})}. $$ Now, let's multiply both sides by \(\Gamma(\mathrm{ix})\) to find \(|\Gamma(\mathrm{ix})|^2\): $$ |\Gamma(\mathrm{ix})|^2 = \Gamma(\mathrm{ix})\Gamma(\mathrm{-ix}). $$ Now, we'll use the reflection formula for the gamma function: $$ \Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}, $$ which holds for complex \(z\) as well. Comparing both expressions, we find that $$ \Gamma(\mathrm{ix})\Gamma(\mathrm{-ix}) = \frac{\pi}{\sin(\pi \mathrm{ix})} . $$ Since \(\sin(\pi \mathrm{ix}) = i\sinh(\pi x)\), we can rewrite the above equation as: $$ |\Gamma(\mathrm{ix})|^2 = \frac{\pi}{x \sinh (\pi x)}, $$ for \(x \in \mathbb{R}_{\neq 0}\). This completes the proof.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Analysis

Complex analysis is a fundamental area in mathematics, dealing with functions that operate on complex numbers. It extends the ideas of calculus to the complex plane, which is made up of numbers that have both a real part and an imaginary part. Complex analysis has a wide range of applications, from solving calculus problems to understanding the flow of electricity.

In relation to the gamma function, complex analysis provides the tools needed to extend this function, which is originally defined for positive real numbers, to the complex plane. This extension allows us to evaluate the gamma function at any complex number (excluding some special cases like the negative real integers and zero), which opens up a myriad of possibilities in both theoretical and applied mathematics.

Complex Conjugate

The complex conjugate of a number is a simple transformation that reflects a point across the real axis on the complex plane. For any complex number z = a + bi (where a and b are real numbers and i is the imaginary unit), the complex conjugate is \bar{z} = a - bi. This process changes the sign of the imaginary part, essentially providing a mirror image of the original number with respect to the real axis.

The relationship between the gamma function and the complex conjugate is a pivotal one. As seen in the provided solution, the gamma function respects the complex conjugate in the sense that \(\Gamma(\bar{z}) = \overline{\Gamma(z)}\), which is a property that plays an essential role in understanding the behavior of the gamma function over the complex plane.

Reflection Formula

The reflection formula is a beautiful symmetry property of the gamma function. It relates \(\Gamma(z)\) and \(\Gamma(1-z)\), providing a bridge between the values of the function at points that are mirrored about the line Re(z) = \frac{1}{2}. Specifically, the formula states:


\[\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}\]


For real numbers, this allows us to evaluate the gamma function at negative inputs using the known values at positive inputs. However, when dealing with complex numbers, the reflection formula can be used alongside the complex conjugate property to derive valuable identities, like the one established in the exercise regarding the modulus squared of the gamma function at purely imaginary arguments.

Sinh Function

The sinh function, also known as the hyperbolic sine, is part of a family of functions analogous to the circular trigonometric functions, but for hyperbolas. For a real number x, it is defined as:


\[\sinh(x) = \frac{e^x - e^{-x}}{2}\]


When x is a real number, sinh(x) measures the distance along a hyperbola, similar to how the sin function measures the distance along a circle. In our context, the sinh function plays an essential role in the solution provided; it appears in the denominator of the equation relating to the modulus squared of the gamma function. By recognizing that the hyperbolic sine is related to the imaginary sine function through \(\sin(ix) = i\sinh(x)\), we can effectively convert trigonometric expressions to hyperbolic ones, further demonstrating the deep connections that exist within mathematics.

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Most popular questions from this chapter

Die Tschebyschev-Polynome k枚nnen 眉ber die Beziehung $$ T_{n}(t)=\cos (n \arccos t) \quad \text { f眉r } t \in[-1,1] $$ definiert werden. Bestimmen Sie \(T_{1}\) und \(T_{2}\) und dr眉cken Sie f眉r \(n \geq 1\) allgemein \(T_{n+1}(t)\) durch \(T_{n}(t)\) und \(T_{n-1}(t)\) aus. (Hinweis: Benutzen Sie die trigonometrischen Identit盲t \(\cos ((n+1) x)=\) \(2 \cos x \cos (n x)-\cos ((n-1) x) .)\)

Zeigen Sie, dass die Koeffizienten von $$ \frac{\mathrm{d}^{n}}{\mathrm{~d} x^{n}}\left(x^{2}-1\right)^{n} $$ die Rekursionsformel $$ a_{k+2}=\frac{k(k+1)-n(n+1)}{(k+2)(k+1)} a_{k} $$ erf眉llen.

Zeigen Sie, dass die Legendre'schen Differenzialgleichung $$ \left(1-x^{2}\right) u^{\prime \prime}-2 x u^{\prime}+\lambda u=0 $$ f眉r den Potenzreihenansatzes \(u(x)=\sum_{k=0}^{\infty} a_{k} x^{k}\) die Rekursionsformel $$ a_{k+2}=\frac{k(k+1)-\lambda}{(k+2)(k+1)} a_{k} $$ liefert.

Welche der folgenden Aussagen sind richtig? 1\. Zylinderfunktionen treten bei Separation als Funktionen des. Abstands \(\rho\) von der \(x_{3}\)-Achse auf. 2\. Zylinderfunktionen sind auf einem Zylinder \(x_{1}^{2}+x_{2}^{2}=\rho_{0}^{2}\) definiert, 3\. Kugelfl盲chenfunktionen treten bei Separation als Funktionen des Abstands \(r\) vom Ursprung \(\mathbf{0}\) auf. 4\. Kugelfal盲chenfunktionen sind auf einer Kugel \(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=\) \(r_{0}^{2}\) definiert.

Zeigen Sie die Beziehung $$ \frac{\Gamma^{\prime}\left(\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)}=\frac{\Gamma^{\prime}(1)}{\Gamma(1)}-2 \ln 2 $$
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