Double integral fail, works on TI

[Albert Chan]

Let y = x sin(t), dy/dt = x cos(t) = √(x² - y²)

\(\large I = \int_0^1 \int _{-x}^x {e^{-x y} \over \sqrt{x^2-y^2}}\;dy\;dx
= \int_0^1 \int _{-\pi/2}^{\pi/2} e^{-x^2 \sin t}\;dt\;dx \)

If w is odd function, taylor series of e^w with odd powers are also odd. Thus, only even powers remained.

\(\large I = \int_0^1 \int _{-\pi/2}^{\pi/2} \cosh(x^2 \sin t)\;dt\;dx
= 2 \int_0^1 \int _0^{\pi/2} \cosh(x^2 \sin t)\;dt\;dx \)

For even power terms, we can use this

Quote:\(\large \int _{-\pi \over 2} ^{\pi \over 2} (\sin x)^{2n} \;dx = \binom{2n}{n}\pi / 2^{2n} \)

\(\large I
= \int_0^1 \sum_{k=0}^{∞} \left({\binom{2k}{k}\pi \over 2^{2k} (2k)!} \right) x^{4k}\;dx
= \sum_{k=0}^{∞} {\pi \over 2^{2k} (k!)^2 (4k+1)}
\) ≈ 3.30423259121