∫_0 ^∞ e^(−x^7 ) sin(x^7 )dx


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
Question Number 124461 by Dwaipayan Shikari last updated on 03/Dec/20

$\int_{0}^{\infty} e^{- x^{7}} s i n (x^{7}) d x$
Commented by Dwaipayan Shikari last updated on 03/Dec/20

$I h a v e f o u n d \frac{Γ (\frac{8}{7}) s i n (\frac{π}{28})}{2^{\frac{1}{14}}}$
Commented by Dwaipayan Shikari last updated on 03/Dec/20

$M y w a y$ $\frac{1}{2 i} \int_{0}^{\infty} e^{- x^{7}} e^{{i x}^{7}} - \frac{1}{2 i} \int_{0}^{\infty} e^{- x^{7} (1 + i)} d x x^{7} = u \Rightarrow 7 x^{6} \Rightarrow \frac{d u}{d x}$ $= \frac{e^{- \frac{π}{2} i}}{2} \int_{0}^{\infty} u^{- \frac{6}{7}} e^{- u (1 - i)} d u - \frac{e^{- \frac{π}{2} i}}{2} \int_{0}^{\infty} u^{- \frac{6}{7}} e^{- u (1 + i)}$ $= Γ (\frac{8}{7}) (\frac{e^{- \frac{π}{2} i} {(1 - i)}^{- \frac{1}{7}}}{2} - \frac{e^{\frac{- π}{2} i} {(1 + i)}^{- \frac{1}{7}}}{2})$ $= \frac{Γ (\frac{8}{7})}{\sqrt[14]{2^{15}}} (e^{- \frac{π}{2} i + \frac{π}{28} i} - e^{- \frac{π}{2} i - \frac{π}{28} i}) = \frac{s i n (\frac{π}{28}) Γ (\frac{8}{7})}{\sqrt[14]{2}}$
	Answered by mnjuly1970 last updated on 03/Dec/20
	$x^7 =t (1/7)∫_0 ^( ∞) e^(−t) sin(t)(dt/t^(6/7) ) I=((−1)/7)Im(∫_0 ^( ∞) e^(−t) e^(−it) (dt/t^(6/7) )) =−(1/7)Im(∫_(0 ) ^( ∞) e^(−t(1+i)) (dt/t^(6/7) )) =^(t(1+i)=u) −(1/7)Im((1/(1+i))∫_0 ^( ∞) e^(−u) (du/((1+i)^(−(6/7)) u^(6/7) ))) −(1/7)Im(1/((1+i)^(1/7) ))∫_0 ^( ∞) e^(−u) u^(((−6)/7)+1−1) =((−1)/7)Im(1/((1+2i−1)^(1/(14)) ))Γ((1/7)) =((−1)/( (2)^(1/(14)) ))Γ((8/7))Im(1/((e^((iπ)/2) )^(1/(14)) )) =((−1)/( (2)^(1/(14)) ))Γ((8/7))Im{cos((π/(28)))−isin((π/(28))) ∴ Ω=∫_0 ^( ∞) e^(−x^7 ) sin(x^7 )dx=(1/( (2)^(1/(14)) ))Γ((8/7))sin((π/(28))) ✓ .....$
	$x^{7} = t$ $\frac{1}{7} \int_{0}^{\infty} e^{- t} s i n (t) \frac{d t}{t^{\frac{6}{7}}}$ $I = \frac{- 1}{7} I m (\int_{0}^{\infty} e^{- t} e^{- i t} \frac{d t}{t^{\frac{6}{7}}})$ $= - \frac{1}{7} I m (\int_{0}^{\infty} e^{- t (1 + i)} \frac{d t}{t^{\frac{6}{7}}})$ $\overset{t (1 + i) = u}{=} - \frac{1}{7} I m (\frac{1}{1 + i} \int_{0}^{\infty} e^{- u} \frac{d u}{{(1 + i)}^{- \frac{6}{7}} u^{\frac{6}{7}}})$ $- \frac{1}{7} I m \frac{1}{{(1 + i)}^{\frac{1}{7}}} \int_{0}^{\infty} e^{- u} u^{\frac{- 6}{7} + 1 - 1}$ $= \frac{- 1}{7} I m \frac{1}{{(1 + 2 i - 1)}^{\frac{1}{14}}} Γ (\frac{1}{7})$ $= \frac{- 1}{\sqrt[14]{2}} Γ (\frac{8}{7}) I m \frac{1}{{(e^{\frac{i π}{2}})}^{\frac{1}{14}}}$ $= \frac{- 1}{\sqrt[14]{2}} Γ (\frac{8}{7}) I m {c o s (\frac{π}{28}) - i s i n (\frac{π}{28})$ $∴ Ω = \int_{0}^{\infty} e^{- x^{7}} s i n (x^{7}) d x = \frac{1}{\sqrt[14]{2}} Γ (\frac{8}{7}) s i n (\frac{π}{28}) ✓$ $. . . . .$
	Commented by Dwaipayan Shikari last updated on 03/Dec/20

	$T h a n k i n g y o u$
	Commented by mnjuly1970 last updated on 03/Dec/20

	$g r a t e f u l$ $y o u r q u e s t i o n s a r e v e r y s t u n n i n g . .$
	Answered by mathmax by abdo last updated on 03/Dec/20
	$let A = ∫_0 ^∞ e^(−x^7 ) sin(x^7 )dx ⇒A=_(x^7 =t) (1/7)∫_0 ^∞ e^(−t) sin(t) t^((1/7)−1) dt =(1/7)∫_0 ^∞ t^(−(6/7)) e^(−t) sint dt =(1/7)Im(∫_0 ^∞ t^(−(6/7)) e^(−t+it) dt)but ∫_0 ^∞ t^(−(6/7)) e^(−(1−i)t) dt =_((1−i)t=u) ∫_0 ^∞ (u^(−(6/7)) /((1−i)^(−(6/7)) )) e^(−u) (du/(1−i)) =(1/((1−i)^(1−(6/7)) ))∫_0 ^∞ u^(−(6/7)) e^(−u) du =(1−i)^(−(1/7)) ∫_0 ^∞ u^((1/7)−1) e^(−u) du =((√2)e^(−((iπ)/4)) )^(−(1/7)) ×Γ((1/7)) =(1/(((√2))^(1/7) )) e^((iπ)/(28)) ×Γ((1/7)) =(1/2^(1/(14)) )×Γ((1/7)){cos((π/(28)))+isin((π/(28)))} ⇒ ∫_0 ^∞ e^(−x^7 ) sin(x^7 )dx =((Γ((1/7)))/7)×((sin((π/(28))))/2^(1/(14)) )$
	$let A = \int_{0}^{\infty} e^{- x^{7}} \sin (x^{7}) dx \Rightarrow A =_{x^{7} = t} \frac{1}{7} \int_{0}^{\infty} e^{- t} \sin (t) t^{\frac{1}{7} - 1} dt$ $= \frac{1}{7} \int_{0}^{\infty} t^{- \frac{6}{7}} e^{- t} sint dt = \frac{1}{7} Im (\int_{0}^{\infty} t^{- \frac{6}{7}} e^{- t + it} dt) but$ $\int_{0}^{\infty} t^{- \frac{6}{7}} e^{- (1 - i) t} dt =_{(1 - i) t = u} \int_{0}^{\infty} \frac{u^{- \frac{6}{7}}}{{(1 - i)}^{- \frac{6}{7}}} e^{- u} \frac{du}{1 - i}$ $= \frac{1}{{(1 - i)}^{1 - \frac{6}{7}}} \int_{0}^{\infty} u^{- \frac{6}{7}} e^{- u} du = {(1 - i)}^{- \frac{1}{7}} \int_{0}^{\infty} u^{\frac{1}{7} - 1} e^{- u} du$ $= {(\sqrt{2} e^{- \frac{i π}{4}})}^{- \frac{1}{7}} \times Γ (\frac{1}{7}) = \frac{1}{{(\sqrt{2})}^{\frac{1}{7}}} e^{\frac{i π}{28}} \times Γ (\frac{1}{7})$ $= \frac{1}{2^{\frac{1}{14}}} \times Γ (\frac{1}{7}) {\cos (\frac{π}{28}) + isin (\frac{π}{28})} \Rightarrow$ $\int_{0}^{\infty} e^{- x^{7}} \sin (x^{7}) dx = \frac{Γ (\frac{1}{7})}{7} \times \frac{\sin (\frac{π}{28})}{2^{\frac{1}{14}}}$
	Commented by mathmax by abdo last updated on 03/Dec/20

	$\frac{1}{7} Γ (\frac{1}{7}) = Γ (1 + \frac{1}{7}) = Γ (\frac{8}{7})$
	Commented by Dwaipayan Shikari last updated on 03/Dec/20

	$T h a n k i n g y o u$
	Commented by mathmax by abdo last updated on 03/Dec/20

	$you are welcome$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum