find the value of integral ∫_0 ^∞ e^(−px) ((sin(qx))/(√x))dx with p>0 and q>0


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Question Number 35228 by abdo mathsup 649 cc last updated on 16/May/18

$f i n d t h e v a l u e o f i n t e g r a l$ $\int_{0}^{\infty} e^{- p x} \frac{s i n (q x)}{\sqrt{x}} d x w i t h p > 0 a n d q > 0$
Commented by abdo mathsup 649 cc last updated on 18/May/18

$l e t p u t I = \int_{0}^{\infty} e^{- p x} \frac{s i n (q x)}{\sqrt{x}} d x c h a n b g e m e n t$ $\sqrt{x} = t g i v e$ $I = \int_{0}^{\infty} e^{- {p t}^{2}} \frac{s i n ({q t}^{2})}{t} 2 t d t$ $I = 2 \int_{0}^{\infty} e^{- {p t}^{2}} s i n ({q t}^{2}) d t$ $= \int_{- \infty}^{+ \infty} e^{- {p t}^{2}} s i n ({q t}^{2}) d t = I m (\int_{- \infty}^{+ \infty} e^{- {p t}^{2} + {i q t}^{2}} d t)$ $b u t \int_{- \infty}^{+ \infty} e^{- {p t}^{2} + {i q t}^{2}} d t = \int_{- \infty}^{+ \infty} e^{- (p - i q) t^{2}} d t$ $=_{\sqrt{p - i q} t = u} \int_{- \infty}^{+ \infty} e^{- u^{2}} \frac{d u}{\sqrt{p - i q}}$ $= \frac{\sqrt{π}}{\sqrt{p - i q}}$ $I = \frac{\sqrt{π}}{\sqrt{p - i q}}$
Commented by abdo mathsup 649 cc last updated on 18/May/18

$I = I m (\frac{\sqrt{π}}{\sqrt{p - i q}}) b u t$ $p - i q = \sqrt{p^{2} + q^{2}} (\frac{p}{\sqrt{p^{2} + q^{2}}} - i \frac{q}{\sqrt{p^{2} + q^{2}}}) = r e^{i θ} \Rightarrow$ $r = \sqrt{p^{2} + q^{2}} a n d t a n θ = - \frac{q}{p} \Rightarrow θ = - a r c t a n (\frac{q}{p})$ $\sqrt{p - i q} = \sqrt{r} e^{i \frac{θ}{2}} = {(p^{2} + q^{2})}^{\frac{1}{4}} e^{- \frac{i}{2} a r c t a n (\frac{q}{p})}$ $I = I m (\sqrt{π} {(p^{2} + q^{2})}^{- \frac{1}{4}} e^{\frac{i}{2} a r c t a n (\frac{q}{p})})$ $I = \sqrt{π} {(p^{2} + q^{2})}^{- \frac{1}{4}} s i n (\frac{1}{2} a r c t a n (\frac{q}{p}))$
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