Chapt_Appendix II - Appendices Appendix II Gamma Function...

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Appendices Appendix II Gamma Function 1. (a) Γ(5) = Γ(4 + 1) = 4! = 24 (b) Γ(7) = Γ(6 + 1) = 6! = 720 (c) Using Example 1 in the text, 2 π µ 1 2 µ 3 2 +1 = 3 2 Γ µ 3 2 . Thus, Γ( 3 / 2) = 4 π/ 3. (d) Using (c) 4 π 3 µ 3 2 µ 5 2 +1 = 5 2 Γ µ 5 2 . Thus Γ( 5 / 2) = 8 π/ 15 2. If t = x 5 , then dt =5 x 4 dx and x 5 dx = 1 5 t 1 / 5 dt .Now Z 0 x 5 e x 5 dx = Z 0 1 5 t 1 / 5 e t dt = 1 5 Z 0 t 1 / 5 e t dt = 1 5 Γ µ 6 5 = 1 5 (0 . 92) = 0 . 184 . 3. If t = x 3 , then dt =3 x 2 dx and x 4 dx = 1 3 t 2 / 3 dt .Now Z 0 x 4 e x 3 dx = Z 0 1 3 t 2 / 3 e t dt = 1 3 Z 0 t 2 / 3 e t dt = 1 3 Γ µ 5 3 = 1 3 (0 . 89) 0 . 297 . 4. If t = ln x =ln 1 x then dt = 1 x dx . Also e t = 1 x ,so x = e t and dx = xdt = e t dt .Thu s Z 1 0 x 3 µ ln 1 x 3 dx = Z 0 ( e t ) 3 t 3 ( e t ) dt = Z 0 t 3 e 4 t dt = Z 0 µ 1 4 u 3 e u µ 1 4 du u =4 t = 1 256 Z 0 u 3 e u du = 1 256 Γ(4) = 1 256 (3!) = 3 128 . 5. Since e t e 1 for 0 t 1, Γ( x )= Z 0 t x 1 e t dt > Z 1 0 t x
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Chapt_Appendix II - Appendices Appendix II Gamma Function...

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