calculate ∫_0 ^∞ x e^(−(x^2 /a^2 )) sin(bx)dx with a>0 and b>0


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Question Number 63214 by mathmax by abdo last updated on 30/Jun/19

$c a l c u l a t e \int_{0}^{\infty} x e^{- \frac{x^{2}}{a^{2}}} s i n (b x) d x w i t h a > 0 a n d b > 0$
Commented bymathmax by abdo last updated on 01/Jul/19
$let I =∫_0 ^∞ x e^(−(x^2 /a^2 )) s in(bx)dx ⇒ 2I =∫_(−∞) ^(+∞) x e^(−(x^2 /a^2 )) sin(bx)dx by parts u^′ =x e^(−(x^2 /a^2 )) and v =sin(bx) ⇒ 2I =[−(a^2 /2) e^(−(x^2 /a^2 )) sin(bx)]_(−∞) ^(+∞) −∫_(−∞) ^(+∞) −(a^2 /2) e^(−(x^2 /a^2 )) b cos(bx) dx =((a^2 b)/2) ∫_(−∞) ^(+∞) e^(−(x^2 /a^2 )) cos(bx)dx let determine A_λ =∫_(−∞) ^(+∞) e^(−λx^2 ) cos(bx)dx with λ>0 A_λ =Re (∫_(−∞) ^(+∞) e^(−λx^2 +ibx) dx) ∫_(−∞) ^(+∞) e^(−λx^2 +ibx) dx =∫_(−∞) ^(+∞) e^(−{ ((√λ)x)^2 −2(√λ)x((ib)/(2(√λ))) −(b^2 /(4λ)) +(b^2 /(4λ))}) dx =∫_(−∞) ^(+∞) e^(−( (√λ)x +((ib)/(2(√λ))))^2 ) e^(−(b^2 /(4λ))) dx =_((√λ)x +((ib)/(2(√υ)))=u) e^(−(b^2 /(4λ))) ∫_(−∞) ^(+∞) e^(−u^2 ) (du/(√λ)) =((√π)/(√λ)) ⇒ A_λ =((√π)/(√λ)) e^(−(b^2 /(4λ))) ⇒ 2I =((a^2 b)/2) ((√π)/(√λ)) e^(−(b^2 /(4λ))) and λ =(1/a^2 ) ⇒ I =((a^2 b)/4) ((√π)/(1/a)) e^(−(b^2 /(4(1/a^2 )))) =((√π)/4) a^3 b e^(−((a^2 b^2 )/4))$
$l e t I = \int_{0}^{\infty} x e^{- \frac{x^{2}}{a^{2}}} s i n (b x) d x \Rightarrow 2 I = \int_{- \infty}^{+ \infty} x e^{- \frac{x^{2}}{a^{2}}} s i n (b x) d x b y p a r t s$ $u^{^{'}} = x e^{- \frac{x^{2}}{a^{2}}} a n d v = s i n (b x) \Rightarrow$ $2 I = {[- \frac{a^{2}}{2} e^{- \frac{x^{2}}{a^{2}}} s i n (b x)]}_{- \infty}^{+ \infty} - \int_{- \infty}^{+ \infty} - \frac{a^{2}}{2} e^{- \frac{x^{2}}{a^{2}}} b c o s (b x) d x$ $= \frac{a^{2} b}{2} \int_{- \infty}^{+ \infty} e^{- \frac{x^{2}}{a^{2}}} c o s (b x) d x l e t d e t e r m i n e A_{λ} = \int_{- \infty}^{+ \infty} e^{- λ x^{2}} c o s (b x) d x w i t h λ > 0$ $A_{λ} = R e (\int_{- \infty}^{+ \infty} e^{- λ x^{2} + i b x} d x)$ $\int_{- \infty}^{+ \infty} e^{- λ x^{2} + i b x} d x = \int_{- \infty}^{+ \infty} e^{- {{(\sqrt{λ} x)}^{2} - 2 \sqrt{λ} x \frac{i b}{2 \sqrt{λ}} - \frac{b^{2}}{4 λ} + \frac{b^{2}}{4 λ}}} d x$ $= \int_{- \infty}^{+ \infty} e^{- {(\sqrt{λ} x + \frac{i b}{2 \sqrt{λ}})}^{2}} e^{- \frac{b^{2}}{4 λ}} d x =_{\sqrt{λ} x + \frac{i b}{2 \sqrt{υ}} = u} e^{- \frac{b^{2}}{4 λ}} \int_{- \infty}^{+ \infty} e^{- u^{2}} \frac{d u}{\sqrt{λ}} = \frac{\sqrt{π}}{\sqrt{λ}} \Rightarrow A_{λ} = \frac{\sqrt{π}}{\sqrt{λ}} e^{- \frac{b^{2}}{4 λ}} \Rightarrow$ $2 I = \frac{a^{2} b}{2} \frac{\sqrt{π}}{\sqrt{λ}} e^{- \frac{b^{2}}{4 λ}} a n d λ = \frac{1}{a^{2}} \Rightarrow$ $I = \frac{a^{2} b}{4} \frac{\sqrt{π}}{\frac{1}{a}} e^{- \frac{b^{2}}{4 \frac{1}{a^{2}}}} = \frac{\sqrt{π}}{4} a^{3} b e^{- \frac{a^{2} b^{2}}{4}}$
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