Looks Like A Gaussian!

Calculus Level 4

If $\displaystyle \int_0^\infty e^{ -x^{\frac {5}{2}} } \ \mathrm{d} x$ can be expressed in the form of $\dfrac {a}{b} \Gamma \left ( \dfrac {2}{5} \right )$ for positive coprime integers $a$ and $b$ . Find $a + b$ .

The answer is 7.

1 solution

Kunal Gupta
Mar 14, 2015

Consider: $I= \displaystyle \int_{0}^{\infty} e^{-x^{n}} dx$
Set $x^{n} =t$ , so $I= \dfrac{1}{n} \displaystyle \int_{0}^{\infty} t^{\frac{1-n}{n}}e^{-t}dt$
Now, Gamma Function is defined as:

$\Gamma(n) = \displaystyle \int_{0}^{\infty} x^{n-1}e^{-x}dx$

So, $I= \dfrac{1}{n} \Gamma \left( \dfrac{1}{n} \right)$ So, the answer becomes : $\dfrac{2}{5} \Gamma \left( \dfrac{2}{5} \right)$

Using the fact that $\Gamma(t + 1) = t \Gamma(t)$ , you can write the answer in different forms. For example, $\frac{2}{5} \Gamma \left( \frac{2}{5} \right) = \frac{1}{1} \Gamma \left( \frac{7}{5} \right) = \frac{5}{7} \Gamma \left(\frac{12}{5} \right).$

Jon Haussmann - 6 years, 3 months ago

Thanks. I have updated the problem statement.

Calvin Lin Staff - 4 years, 7 months ago

Taking Jon Haussmann comment into consideration I think you should edit the statement of the problem

Arkin Dharawat - 5 years, 7 months ago

Looks Like A Gaussian!

The answer is 7.

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