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Question Number 46091 by sandeepkeshari0797@gmail.com last updated on 21/Oct/18

Commented by maxmathsup by imad last updated on 21/Oct/18
$let f(t)=∫_0 ^u ((sinx)/x)e^(−tx) dx with t≥o we have f^′ (t)=−∫_0 ^u sinx e^(−tx) dx ⇒−f^′ (t)=Im(∫_0 ^u e^(ix) e^(−tx) dx)=Im ( ∫_0 ^u e^((i−t)x) dx) but ∫_0 ^u e^((i−t)x) dx =[(1/(i−t)) e^((i−t)x) ]_0 ^u =(1/(i−t))( e^((i−t)u) −1) = (1/(t−i))( 1−e^(−tu) (cosu +isinu))=((t+i)/(t^2 +1)){ 1−e^(−tu) cosu −i e^(−tu) sinu} =((t−t e^(−tu) cosu−it e^(−tu) sinu +i(1−e^(−tu) cosu )+e^(−tu) sinu)/(1+t^2 )) =((t−t e^(−tu) cosu +e^(−tu) sinu +i(1−e^(−tu) cosu−t e^(−tu) sinu))/(1+t^2 )) ⇒ f^′ (t)=((t e^(−tu) sinu +e^(−tu) cosu −1)/(1+t^2 )) ⇒ f(t)= ∫_0 ^t ((x e^(−xu) sinu +e^(−xu) cosu−1)/(1+x^2 ))dx+c c=f(o)=∫_0 ^u ((sinx)/x)dx ....be continued...$
$l e t f (t) = \int_{0}^{u} \frac{s i n x}{x} e^{- t x} d x w i t h t ⩾ o w e h a v e f^{^{'}} (t) = - \int_{0}^{u} s i n x e^{- t x} d x$ $\Rightarrow - f^{^{'}} (t) = I m (\int_{0}^{u} e^{i x} e^{- t x} d x) = I m (\int_{0}^{u} e^{(i - t) x} d x) b u t$ $\int_{0}^{u} e^{(i - t) x} d x = {[\frac{1}{i - t} e^{(i - t) x}]}_{0}^{u} = \frac{1}{i - t} (e^{(i - t) u} - 1)$ $= \frac{1}{t - i} (1 - e^{- t u} (c o s u + i s i n u)) = \frac{t + i}{t^{2} + 1} {1 - e^{- t u} c o s u - i e^{- t u} s i n u}$ $= \frac{t - t e^{- t u} c o s u - i t e^{- t u} s i n u + i (1 - e^{- t u} c o s u) + e^{- t u} s i n u}{1 + t^{2}}$ $= \frac{t - t e^{- t u} c o s u + e^{- t u} s i n u + i (1 - e^{- t u} c o s u - t e^{- t u} s i n u)}{1 + t^{2}} \Rightarrow$ $f^{^{'}} (t) = \frac{t e^{- t u} s i n u + e^{- t u} c o s u - 1}{1 + t^{2}} \Rightarrow f (t) = \int_{0}^{t} \frac{x e^{- x u} s i n u + e^{- x u} c o s u - 1}{1 + x^{2}} d x + c$ $c = f (o) = \int_{0}^{u} \frac{s i n x}{x} d x . . . . b e c o n t i n u e d . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 21/Oct/18

	$s i n x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + . . .$ $\int \frac{s i n x}{x} d x$ $\int 1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} - \frac{x^{6}}{7!} + . . . d x$ $= x - \frac{x^{3}}{3 \times 3!} + \frac{x^{5}}{5 \times 5!} - \frac{x^{7}}{7 \times 7!} + . . . C$
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