Reflection relation for gamma function - Revision history

Vipul: 5 revisions

2008-05-18T19:17:54Z

5 revisions

← Older revision	Revision as of 19:17, 18 May 2008
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2008-05-01T20:39:16Z

← Older revision	Revision as of 20:39, 1 May 2008
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2008-05-01T19:08:49Z

Facts used

← Older revision		Revision as of 19:08, 1 May 2008
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	* The following identity, for <math>0 < \operatorname{Re}(a) < 1</math>:		* The following identity, for <math>0 < \operatorname{Re}(a) < 1</math>:

	<math>\int_0^1 \frac{v^{a-1}}{1 + v} = \frac{\pi}{\sin \pi z}</math>		<math>\int_0^1 \frac{v^{a-1}}{1 + v} = \frac{\pi}{\sin \pi a}</math>

	==Proof==		==Proof==

2008-05-01T19:08:36Z

Facts used

← Older revision		Revision as of 19:08, 1 May 2008
Line 10:		Line 10:
	* The following identity, for <math>0 < \operatorname{Re}(a) < 1</math>:		* The following identity, for <math>0 < \operatorname{Re}(a) < 1</math>:

	<math>\int_0^1 \frac{v^{a-1}}{1 + v} = \frac{\pi{\sin \pi z}</math>		<math>\int_0^1 \frac{v^{a-1}}{1 + v} = \frac{\pi}{\sin \pi z}</math>

	==Proof==		==Proof==

2008-05-01T19:08:18Z

Facts used

← Older revision		Revision as of 19:08, 1 May 2008
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	* [[Recurrence relation for gamma function]]: This shows that it suffices to prove the result for <math>\operatorname{Re}(z) \in (0,1)</math>		* [[Recurrence relation for gamma function]]: This shows that it suffices to prove the result for <math>\operatorname{Re}(z) \in (0,1)</math>
			* The following identity, for <math>0 < \operatorname{Re}(a) < 1</math>:

			<math>\int_0^1 \frac{v^{a-1}}{1 + v} = \frac{\pi{\sin \pi z}</math>

			==Proof==

			===Proof of the integration identity===

			Pick a branch of the logarithm that is slit along the positive real axis, and thus rewrite:

			<math>v^{a - 1} = e^{(a - 1)\log v}</math>

			Where the branch of logarithm is chosen so that the approach to the positive real axis from the upper half-plane side is the usual logarithm.

			Next, apply the [[keyhole contour integration method]] to compute this integral. The inner and outer circle integrals approach zero, and if we define:

			<math>I ::= \int_0^1 \frac{v^{a-1}}{1 + v}</math>

			Then, we'll obtain:

			<math>I(e^{2\pi i (a - 1)} - 1) = 2\pi i \operatorname{res}(\frac{e^{(a - 1)\log v}}{1 + v}; -1)</math>

			The residue at <math>-1</math> is simple <math>e^{\pi i (a - 1)}</math> and simplifying the expression, we obtain the required expression for <math>I</math>.

			===Proving the result using the integration identity===