$\gamma,\Gamma$ -day Problem 2

We will use $\Gamma(x)=\int_0^{\infty}t^{x-1}e^{-t}\,dt$ $\Gamma(\frac{1}{2}+n)=\frac{(2n-1)!!}{2^n}\sqrt{\pi}$ & we will first solve the general case $I=\int_0^{\infty}x^me^{-kx^n}\,dx$ take $y=kn^x$ $\Rightarrow I=\frac{1}{n\cdot k^{\frac{m+1}{n}}}\int_0^{\infty}y^{\frac{m+1}{n}-1}e^{-y}\,dy$ $I=\frac{1}{n\cdot k^{\frac{m+1}{n}}}\Gamma(\frac{m+1}{n})$ the integral in the question can be written as $\int_0^{\infty}x^6e^{-3x^2}\,dx$ with m=6 , k=3 , n=2 as compared to above . So wwe get that integral is equal to $\frac{1}{2\cdot3^{\frac{7}{2}}}\Gamma(\frac{7}{2})$ On simplifying which is equal to $\frac{5\sqrt{\pi}}{2^4\cdot3^{\frac{5}{2}}}$ Hence we get our answer a+b+c+d =15

We can also solve this problem without using the gamma function. First, we can write the integral above as

$\begin{aligned} \int_{0}^{\infty} x^6 e^{-3x^2} dx. \end{aligned}$

With the substitution $y = 3x^2,$ (I'm using y because I'm going to do integration by parts later) we see that the integral is equal to

$\begin{aligned} \frac{1}{6 \cdot 3^{5/2}} \int_{0}^{\infty} y^{5/2} e^{-y} dy, \end{aligned}$

where the $y$ is to the power of $5/2$ because of the y-substitution. From here, we can do integration by parts with $u = y^{5/2}$ , $dv = e^{-y} dy$ to see that this integral (without the constant out front) amounts to

$\begin{aligned} -y^{5/2} e^{-y} + \frac{5}{2} \int e^{-y} y^{3/2} dy. \end{aligned}$

We can see that the first term is going to go to 0 at both limits as the exponential function dominates as x (and y) approach infinity, and at 0 x (and y) will also yield a value of 0. Doing integration by parts again and seeing that the first term will go to 0 again (including that weird constant from the beginning), we get that this amounts to

$\begin{aligned} \frac{1}{6 \cdot 3^{5/2}} \cdot \frac{3}{2} \cdot \frac{5}{2} \int_{0}^{\infty} \sqrt{y} e^{-y} dy. \end{aligned}$

From here, we can do another u-substitution with $u = \sqrt{y}$ , and then do integration by parts to see that this last integral equals $\frac{\sqrt{\pi}}{2}$ . This means our final answer is

$\begin{aligned} \frac{5\sqrt{\pi}}{2^4 \cdot 3^{5/2}}, \end{aligned}$

so $a=5, b=2, c=3, d=5,$ and our answer is $5+2+3+5=15.$