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Question Number 153430 by mnjuly1970 last updated on 07/Sep/21

Commented by mnjuly1970 last updated on 07/Sep/21

$t h a n k s a l o t . . . .$
Commented by mindispower last updated on 08/Sep/21

$p l e a s u r$
	Answered by mindispower last updated on 07/Sep/21

	$J = - \frac{1}{2} \int_{0}^{\infty} \frac{e^{- x} s i n (x) l n (x)}{x} d x$ $f (a) = \int_{0}^{\infty} e^{- x (1 + i)} x^{a} d x$ $\int_{D} e^{- z} z^{a} d z, D_{R} = [0, R] \cup [{R e}^{i θ}, θ \in [0, \frac{π}{4}]] \cup [x (1 + i), x \in [0, \frac{R}{\sqrt{2}}]]$ $\int_{D_{R}} e^{- z} z^{a} = 0 = \int_{0}^{R} e^{- z} z^{a} d z + \int_{C R} f (z) d z + {(1 + i)}^{a + 1} \int_{\frac{R}{\sqrt{2}}}^{0} e^{- x (1 + i)} x^{a} = 0$ $\lim_{R \to \infty} \int_{C_{R}} f (z) d z = 0 \Rightarrow$ $\Rightarrow (1 \int_{0}^{\infty} e^{- x (1 + i)} x^{a} d x = \frac{1}{{(1 + i)}^{a + 1}} Γ (1 + a))$ $f (a) = (2^{\frac{- a - 1}{2}}) e^{- \frac{i π (a + 1)}{4}} Γ (1 + a))$ $I m f (a) = - \frac{1}{2^{\frac{1 + a}{2}}} s i n (\frac{a + 1}{4} π) Γ (1 + a)$ $J = - \frac{1}{2} . \frac{d i m f (a)}{d a} ∣_{a = - 1}$ $= \frac{1}{2} (- \frac{1}{2} (l n (2) 2^{- \frac{1 + a}{2}} s i n (\frac{a + 1}{4} π) Γ (1 + a) + \frac{π}{4} c o s (\frac{a + 1}{4} π) Γ (1 + a) 2^{- \frac{1 + a}{2}}$ $+ Ψ (1 + a) Γ (1 + a) s i n (\frac{a + 1}{4} π) . 2^{- \frac{1 + a}{2} π})$ $u s i n g Ψ (1 + a) = Ψ (2 + a) - \frac{1}{a + 1})$ $c o s (\frac{a + 1}{4} π) = 1 + o (a + 1), a \to - 1$ $s i n (\frac{a + 1}{4} π) = \frac{a + 1}{4} π + o (a + 1)$ $w e g e t$ $J = - \frac{1}{2} (- \frac{1}{2} l n (2) . \frac{π}{4} \lim_{a \to - 1} (1 + a) Γ (1 + a) + \frac{π}{4} Γ (1 + a)$ $(Ψ (2 + a) - \frac{1}{a + 1}) Γ (1 + a) . \frac{π}{4} (1 + a)))$ $= \frac{l n (2) π}{16} + \lim_{a \to - 1} \frac{π}{4} (Γ (a + 1) + Ψ (2 + a) Γ (2 + a) - \frac{Γ (2 + a)}{1 + a})$ $= \frac{π}{16} l n (2) \frac{π}{4} - \frac{π}{8} \lim_{a \to - 1} (\frac{(1 + a) Γ (1 + a) - Γ (2 + a)}{1 + a} + Ψ (1))$ $Γ (2 + a) = Γ (1 + a) . (1 + a)$ $w e g e t \frac{π}{8} (\frac{l n (2)}{2} + γ) = \frac{π}{8} (l n (\sqrt{2}) + γ)$
	Commented by puissant last updated on 07/Sep/21

	$S i r m i n d i s p o w e r y o u a r e a b o s s . .$
	Commented by mnjuly1970 last updated on 08/Sep/21

	$t h a n k y o u s o m u c h m r p o w e r$
	Commented by mindispower last updated on 08/Sep/21

	$w i t h e p l e a s u r h a v e a n i c e d a y$
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