Good afternoon to everybody
Could you please help to get a closed form of the integral above
where a is a positive real number.
A form based on hypergeometric functions will be also considered as closed form
Best regards
2026-05-16 11:15:05.1778930105
Closed form for the integral \[\int_{0}^{\infty}e^{-ax}e^{-x^{4}}dx\]
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Using Wolfram Alpha, I get...
$$\int_0^{\infty }e^{-a x} e^{-x^4} \, dx$$ $$= \Gamma \left(\frac{5}{4}\right) \, _0F_2\left(;\frac{1}{2},\frac{3}{4};\frac{a^4}{256}\right)-\frac{1}{24} a \left(a \left(a \, _1F_3\left(1;\frac{5}{4},\frac{3}{2},\frac{7}{4};\frac{a^4}{256}\right)-3 \Gamma \left(\frac{3}{4}\right) \, _0F_2\left(;\frac{5}{4},\frac{3}{2};\frac{a^4}{256}\right)\right)+6 \sqrt{\pi } \, _0F_2\left(;\frac{3}{4},\frac{5}{4};\frac{a^4}{256}\right)\right)$$
which is not pretty...