1) calculate I_n = ∫_0 ^∞ x^n e^((1−i)x) dx with n integr natural and i^2 =−1 2) find ∫


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Question Number 46612 by maxmathsup by imad last updated on 29/Oct/18

$1) c a l c u l a t e I_{n} = \int_{0}^{\infty} x^{n} e^{(1 - i) x} d x w i t h n i n t e g r n a t u r a l a n d i^{2} = - 1$ $2) f i n d \int_{0}^{\infty} x^{4 k + 3} x s i n x d x .$
Commented by maxmathsup by imad last updated on 29/Oct/18

$2) t h e Q i s f i n d \int_{0}^{\infty} e^{- x} x^{4 k + 3} s i n x d x .$
Commented by maxmathsup by imad last updated on 29/Oct/18

$1) t h e Q i s {c a l c u l a t e I}_{n} = \int_{0}^{\infty} x^{n} e^{(- 1 + i) x} d x$
Commented by maxmathsup by imad last updated on 29/Oct/18
$1) we have I_n = ∫_0 ^∞ x^n e^((−1+i)x) dx =∫_0 ^∞ x^n e^(−(1−i)x) dx changement (1−i)x=t give I_n =∫_0 ^∞ ((t/(1−i)))^n e^(−t) (dt/(1−i)) =(1/((1−i)^(n+1) )) ∫_0 ^∞ t^n e^(−t) dt =(1/((1−i)^(n+1) )) Γ(n+1) =((n!)/((1−i)^(n+1) )) =((n!)/(((√2)e^(−((iπ)/4)) )^(n+1) )) =((n!)/2^((n+1)/2) ) e^(i(((n+1)/4)π)) =((n!)/2^((n+1)/2) ){ cos((((n+1)π)/4))+i sin((((n+1)π)/4))}$
$1) w e h a v e I_{n} = \int_{0}^{\infty} x^{n} e^{(- 1 + i) x} d x = \int_{0}^{\infty} x^{n} e^{- (1 - i) x} d x c h a n g e m e n t (1 - i) x = t$ $g i v e I_{n} = \int_{0}^{\infty} {(\frac{t}{1 - i})}^{n} e^{- t} \frac{d t}{1 - i} = \frac{1}{{(1 - i)}^{n + 1}} \int_{0}^{\infty} t^{n} e^{- t} d t$ $= \frac{1}{{(1 - i)}^{n + 1}} Γ (n + 1) = \frac{n!}{{(1 - i)}^{n + 1}} = \frac{n!}{{(\sqrt{2} e^{- \frac{i π}{4}})}^{n + 1}}$ $= \frac{n!}{2^{\frac{n + 1}{2}}} e^{i (\frac{n + 1}{4} π)} = \frac{n!}{2^{\frac{n + 1}{2}}} {c o s (\frac{(n + 1) π}{4}) + i s i n (\frac{(n + 1) π}{4})}$
Commented by maxmathsup by imad last updated on 29/Oct/18

$2) w e h a v e \int_{0}^{\infty} e^{- x} x^{4 k + 3} s i n x d x = I m (\int_{0}^{\infty} x^{4 k + 3} e^{(- 1 + i) x} d x)$ $= I m (I_{4 k + 3}) = \frac{(4 k + 3)!}{2^{2 k + 2}} s i n (\frac{(4 k + 4) π}{4}) = \frac{(4 k + 3)!}{2^{2 k + 2}} s i n ((k + 1) π) = 0$
	Answered by tanmay.chaudhury50@gmail.com last updated on 29/Oct/18
	$(1−i)x=−t [ t=(i−1)x] dx=((−dt)/((1−i))) ∫_0 ^∞ (t^n /((i−1)^n ))×e^(−t) ×((−dt)/((1−i))) (1/((i−1)^(n+1) ))∫_0 ^∞ e^(−t) t^(n+1−1) dt (1/((i−1)^(n+1) ))×⌈(n+1) (1/((i−1)^(n+1) ))×n! =(((−1)^(n+1) )/((1−i)^(n+1) ))×(((1+i)^(n+1) )/((1+i)^(n+1) ))×n! =(((−1)^(n+1) )/((2)^(n+1) ))×{(√2) ((1/((√2) ))+i×(1/(√2)))}^(n+1) ×n! =(((−1)^(n+1) )/2^((n+1)/2) )×{cos(π/4)+isin(π/4)}^(n+1) ×n! =(((−1)^(n+1) )/2^((n+1)/2) )×{cos(2kπ+(π/4))+isin(2kπ+(π/4))}^(n+1) ×n!$
	$(1 - i) x = - t [t = (i - 1) x]$ $d x = \frac{- d t}{(1 - i)}$ $\int_{0}^{\infty} \frac{t^{n}}{{(i - 1)}^{n}} \times e^{- t} \times \frac{- d t}{(1 - i)}$ $\frac{1}{{(i - 1)}^{n + 1}} \int_{0}^{\infty} e^{- t} t^{n + 1 - 1} d t$ $\frac{1}{{(i - 1)}^{n + 1}} \times ⌈ (n + 1)$ $\frac{1}{{(i - 1)}^{n + 1}} \times n!$ $= \frac{{(- 1)}^{n + 1}}{{(1 - i)}^{n + 1}} \times \frac{{(1 + i)}^{n + 1}}{{(1 + i)}^{n + 1}} \times n!$ $= \frac{{(- 1)}^{n + 1}}{{(2)}^{n + 1}} \times {\sqrt{2} (\frac{1}{\sqrt{2}} + i \times \frac{1}{\sqrt{2}})}^{n + 1} \times n!$ $= \frac{{(- 1)}^{n + 1}}{2^{\frac{n + 1}{2}}} \times {c o s \frac{π}{4} + i s i n \frac{π}{4}}^{n + 1} \times n!$ $= \frac{{(- 1)}^{n + 1}}{2^{\frac{n + 1}{2}}} \times {c o s (2 k π + \frac{π}{4}) + i s i n (2 k π + \frac{π}{4})}^{n + 1} \times n!$
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