calculate-0-x-e-x-2-a-2-sin-bx-dx-with-a-gt-0-and-b-gt-0-

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 by mathmax by abdo last updated on 01/Jul/19

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}}