Question

$A = \int_{0}^{\infty} x^2 e^{-x^4} dx$ and $B = \int_{0}^{\infty} e^{-x^4} dx$. If the value of $AB$ is $\frac{1}{k}$, then $[k]$ is

(Note: $[k]$ is greatest integer less than or equal to k)

Answer 1

3

Answer 2

The problem requires the evaluation of two definite integrals, $A$ and $B$, and then finding the greatest integer less than or equal to $k$, where $AB = \frac{1}{k}$. These integrals can be evaluated using the Gamma function.

The Gamma function is defined as: $\Gamma(z) = \int_{0}^{\infty} t^{z-1} e^{-t} dt$

Let's evaluate integral $B$: $B = \int_{0}^{\infty} e^{-x^4} dx$ To transform this into the Gamma function form, let $t = x^4$. Then $x = t^{1/4}$. Differentiating with respect to $t$, we get $dx = \frac{1}{4} t^{(1/4)-1} dt = \frac{1}{4} t^{-3/4} dt$. When $x=0$, $t=0$. When $x=\infty$, $t=\infty$. Substituting these into the integral for $B$: $B = \int_{0}^{\infty} e^{-t} \left(\frac{1}{4} t^{-3/4}\right) dt$ $B = \frac{1}{4} \int_{0}^{\infty} t^{-3/4} e^{-t} dt$ Comparing this with the definition of the Gamma function, we have $z-1 = -3/4$, which implies $z = 1 - 3/4 = 1/4$. So, $B = \frac{1}{4} \Gamma\left(\frac{1}{4}\right)$

Now, let's evaluate integral $A$: $A = \int_{0}^{\infty} x^2 e^{-x^4} dx$ Using the same substitution $t = x^4$, we have $x = t^{1/4}$ and $dx = \frac{1}{4} t^{-3/4} dt$. Also, $x^2 = (t^{1/4})^2 = t^{1/2}$. Substituting these into the integral for $A$: $A = \int_{0}^{\infty} t^{1/2} e^{-t} \left(\frac{1}{4} t^{-3/4}\right) dt$ $A = \frac{1}{4} \int_{0}^{\infty} t^{(1/2) - (3/4)} e^{-t} dt$ $A = \frac{1}{4} \int_{0}^{\infty} t^{2/4 - 3/4} e^{-t} dt$ $A = \frac{1}{4} \int_{0}^{\infty} t^{-1/4} e^{-t} dt$ Comparing this with the definition of the Gamma function, we have $z-1 = -1/4$, which implies $z = 1 - 1/4 = 3/4$. So, $A = \frac{1}{4} \Gamma\left(\frac{3}{4}\right)$

Next, we need to find the product $AB$: $AB = \left(\frac{1}{4} \Gamma\left(\frac{3}{4}\right)\right) \left(\frac{1}{4} \Gamma\left(\frac{1}{4}\right)\right)$ $AB = \frac{1}{16} \Gamma\left(\frac{1}{4}\right) \Gamma\left(\frac{3}{4}\right)$ We use the Gamma function reflection formula, which states: $\Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin(\pi z)}$ Let $z = 1/4$. Then $1-z = 3/4$. $\Gamma\left(\frac{1}{4}\right) \Gamma\left(1-\frac{1}{4}\right) = \Gamma\left(\frac{1}{4}\right) \Gamma\left(\frac{3}{4}\right) = \frac{\pi}{\sin\left(\frac{\pi}{4}\right)}$ Since $\sin(\pi/4) = \frac{1}{\sqrt{2}}$, we have: $\Gamma\left(\frac{1}{4}\right) \Gamma\left(\frac{3}{4}\right) = \frac{\pi}{1/\sqrt{2}} = \pi\sqrt{2}$ Substitute this back into the expression for $AB$: $AB = \frac{1}{16} (\pi\sqrt{2}) = \frac{\pi\sqrt{2}}{16}$ We are given that $AB = \frac{1}{k}$. So, $\frac{1}{k} = \frac{\pi\sqrt{2}}{16}$ This implies: $k = \frac{16}{\pi\sqrt{2}}$ To find $[k]$, we need to approximate the value of $k$. Using approximations $\pi \approx 3.14159$ and $\sqrt{2} \approx 1.41421$: $k \approx \frac{16}{(3.14159)(1.41421)}$ $k \approx \frac{16}{4.44288}$ $k \approx 3.60119$ The greatest integer less than or equal to $k$ is $[k] = [3.60119] = 3$.