# Difficulty with Asymptotic Expansion of $\int_{0}^{1}\sqrt{t}e^{ixt}dt$

In the book Advanced Mathematical Methods for Scientists and Engineers by Bender and Orszag (question 6.50) we are asked to compute the asymptotic expansion of $\int_{0}^{1}\sqrt{t}e^{ixt}dt$ fully. We are told the answer is

$$\frac{i\sqrt{\pi}}{2x^{\frac{3}{2}}}e^{i\frac{\pi}{4}}-\frac{i}{x\sqrt{\pi}}e^{ix}\left[1+\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{-i}{x}\right)^{n+1}\Gamma\left(n+\frac{1}{2}\right)\right]$$

My problem is that the answer I keep computing is the following:

$$\frac{i\sqrt{\pi}}{2x^{\frac{3}{2}}}e^{i\frac{\pi}{4}}-\frac{i}{x\sqrt{\pi}}e^{ix}\left[\sqrt{\pi}-\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{-i}{x}\right)^{n+1}\Gamma\left(n+\frac{1}{2}\right)\right]$$

I obtain this by using the original expansion up to first order given in the book and then proceeding by using the hint given in the book which says:

$$\int_{0}^{1}\frac{1}{\sqrt{t}}e^{ixt}dt=\int_{0}^{\infty}\frac{1}{\sqrt{t}}e^{ixt}dt-\int_{1}^{\infty}\frac{1}{\sqrt{t}}e^{ixt}dt$$

The expansion started in the book says that:

$$\int_{0}^{1}\sqrt{t}e^{ixt}dt=\frac{-i}{x}e^{ix}+\frac{i}{2x}\int_{0}^{1}\frac{1}{\sqrt{t}}e^{ixt}dt$$

From here I use the hint and perform a contour integration to get the term $\frac{i\sqrt{\pi}}{2x^{\frac{3}{2}}}e^{i\frac{\pi}{4}}$ and end up with:

$$\int_{0}^{1}\sqrt{t}e^{ixt}dt=\frac{-i}{x}e^{ix}+\int_{0}^{1}\sqrt{t}e^{ixt}dt$$

$$=\frac{-i}{x}e^{ix}+\frac{i\sqrt{\pi}}{2x^{\frac{3}{2}}}e^{i\frac{\pi}{4}}-\frac{i}{2x}\int_{1}^{\infty}\frac{1}{\sqrt{t}}e^{ixt}dt$$

$$=\frac{i\sqrt{\pi}}{2x^{\frac{3}{2}}}e^{i\frac{\pi}{4}}-\frac{i}{x}\left(e^{ix}+\frac{1}{2}\int_{1}^{\infty}\frac{1}{\sqrt{t}}e^{ixt}dt\right)$$

Finally I repeatedly use integration by parts to try an obtain the answer. I have obviously made a mistake somewhere. If someone could point it out to me I would be very grateful. Thank you in advance.

• I haven't tried computing the series myself by numerically it appears that your answer yields a much better approximation to the integral than the one given by the book. I suspect there is a typo in the book's answer. – Antonio Vargas Oct 29 '14 at 20:59
• @AntonioVargas Thank you for your response! – user99163 Oct 30 '14 at 3:50