Need help solving this question - I've tried solving it multiple times but to no avail:
$$
I=\int_0^\infty e^{-y^3}.\sqrt y~dy
$$
My approach has been to substitute:
$$
y^3=t\Rightarrow 3.y^2~dy = dt \Rightarrow I= \frac13\int_0^\infty e^{-t}t^{-1/2}~dt
$$
I'm lost on solving the above Integral. A similar problem suggested the answer in terms of e.r.f.
Related Question
However, I'm unable to understand how to solve this integral and get a numerical answer such as $\frac{\sqrt\pi}3 $. (Also, I'm curious as to how did $\pi$ appear in the expression? - My understanding of error functions is very limited. Wikipedia doesn't offer much help in explaining it either).
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Abhin33t_S
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Have you heard about the Gamma function or the gaussian integral? – mickep Apr 10 '17 at 18:25
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No, I haven't used it before. I'll check. – Abhin33t_S Apr 10 '17 at 18:27
2 Answers
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Hint. The gamma function is given by $$ \Gamma(s)=\int_0^\infty t^{s-1}e^{-t}~dt, \quad s>0. $$ Can you apply it here? Do you see the corresponding value for $s$?
Olivier Oloa
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1It's mind-blowing to know $ -{\frac12}!$ becomes $\sqrt\pi$. I now understand why it was a particularly tough problem to solve – Abhin33t_S Apr 10 '17 at 18:45
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@Abhin33t_S I think you mean $\frac12 !=\sqrt{\pi}$, sure this is not trivial... – Olivier Oloa Apr 10 '17 at 18:55
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Try setting $u=y^{\frac{3}{2}}$. Your integral reduces to $$ \int_0^\infty e^{-y^3}\sqrt{y}dy=\frac{2}{3}\int_0^\infty e^{-u^2}du=\frac{1}{3}\sqrt{\pi} $$
operatorerror
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