solving integral

Question

How can I solve this integral to get the result as follow:

$${\sqrt{\alpha} \over 2\pi} \int_{0}^{t} {1\over \sqrt{r^{3}(t-r)}}[\sin({\alpha\over2r})+\cos({\alpha\over2r})] \mathrm{d}r= {1\over{\sqrt{\pi t}}} \cos({\alpha\over2t}) $$ Thanks!

Answer 1

Let $$I(t) = {\sqrt{\alpha} \over 2\pi} \int_{0}^{t} {1\over \sqrt{r^{3}(t-r)}}[\sin({\alpha\over2r})+\cos({\alpha\over2r})] \mathrm{d}r $$ then let $x = (\alpha/2 r)$ to obtain \begin{align} I(t) &= \frac{\alpha \, \sqrt{\alpha}}{4 \pi} \left(\frac{\alpha}{2}\right)^{3/2} \, \int_{\alpha/2t}^{\infty} \left(\sin(x) + \cos(x) \right) \, \frac{dx}{\sqrt{xt - \alpha/2}} \\ &= \frac{1}{\sqrt{2 t} \, \pi} \, \int_{0}^{\infty} \frac{\sin(x) + \cos(x)}{\sqrt{x - \frac{\alpha}{2t}}} \, dx \end{align} Let $x = u^{2} + \frac{\alpha}{2t}$ to obtain \begin{align} I(t) &= \sqrt{\frac{2}{\pi}} \, \frac{1}{\sqrt{\pi \, t}} \, \int_{0}^{\infty} \left[ \sin\left(u^{2} + \frac{\alpha}{2 t} \right) + \cos\left(u^{2} + \frac{\alpha}{2t} \right) \right] \, du \end{align} Now using \begin{align} \int_{0}^{\infty} \sin(u^{2} + a) \, du &= \frac{1}{2} \, \sqrt{\frac{\pi}{2}} \, \left( \sin a + \cos a \right) \\ \int_{0}^{\infty} \cos(u^{2} + a) \, du &= \frac{1}{2} \, \sqrt{\frac{\pi}{2}} \, \left(\cos a - \sin a \right) \end{align} then the desired result becomes \begin{align} I(t) = \frac{1}{\sqrt{\pi \, t}} \, \cos\left(\frac{\alpha}{2 \, t}\right) \end{align} Thus, \begin{align} {\sqrt{\alpha} \over 2\pi} \int_{0}^{t} {1\over \sqrt{r^{3}(t-r)}}[\sin({\alpha\over2r})+\cos({\alpha\over2r})] \mathrm{d}r = \frac{1}{\sqrt{\pi \, t}} \, \cos\left(\frac{\alpha}{2 \, t}\right) \end{align}