A calculus problem by Rahil Sehgal

For Generalizing things a bit let us consider the integral,

$\displaystyle F_n(m) \; = \; \int_0^\infty \int_0^\infty\cdots \int_0^\infty\;\dfrac{dx_1\;dx_2\;\cdots dx_n}{(1+x_1+x_2+\cdots+x_n)^m}$

Noting that, $\displaystyle \dfrac{1}{(1+x_1+x_2+\cdots+x_n)^m}=\dfrac{1}{\Gamma(m)}\left(\int_0^\infty e^{-u(1+\sum_{i} x_i)}u^{m-1}\; du\right)$ and substituting this back into $F_n(m)$ we have,

$\displaystyle \begin{aligned} F_n(m)&=\int_0^\infty \int_0^\infty\cdots \int_0^\infty\; \dfrac{1}{\Gamma(m)}\left(\int_0^\infty e^{-u(1+\sum_{i} x_i)}u^{m-1}\; du\right) \; dx_1\;dx_2\cdots dx_n \\ &= \dfrac{1}{\Gamma(m)} \int_0^\infty e^{-u}\;\left(\int_0^\infty e^{-ux_1}\; dx_1\right)\cdots\left(\int_0^\infty e^{-ux_n}\; dx_n\right)\; u^{m-1}\; du \\ &= \dfrac{1}{\Gamma(m)}\int_0^\infty e^{-u}\;\left(\int_0^\infty e^{-ux}\;dx\right)^n\;u^{m-1}\;du \\ &=\dfrac{1}{\Gamma(m)}\int_0^\infty e^{-u}\times \dfrac{1}{u^n}u^{m-1}\; du \\ &= \dfrac{1}{\Gamma(m)}\int_0^\infty e^{-u}u^{m-n-1}\;du \\ &=\dfrac{\Gamma(m-n)}{\Gamma(m)} \end{aligned}$

Put $n=3,m=7/2$ to derive the answer as $\boxed{\dfrac{8}{15}}$

Let us change the variables from $(x,y,z)$ to $(x,y,z')$ where $z'=x+y+z$ . Then, $dx\, dy\, dz=dx\, dy\, dz'$ and for particular $(x,y)$ , $z'$ varies from $(x+y)$ to $\infty$ . Then, $\begin{aligned}&\quad \int_{0}^{\infty} \displaystyle\int_{0}^{\infty} \displaystyle\int_{0}^{\infty} \dfrac{dx \; dy \; dz}{( 1+x+y+z)^{{7/2}}}\\&=\displaystyle\int_{0}^{\infty} dx \displaystyle\int_{0}^{\infty} dy \displaystyle\int_{(x+y)}^{\infty} \dfrac{ dz'}{( 1+z')^{{7/2}}}\\ &=\frac{2}{5}\int_{0}^{\infty} dx \displaystyle\int_{0}^{\infty} \dfrac{ dy}{( 1+x+y)^{5/2}}\\ &=\frac{2}{5}\int_{0}^{\infty} dx \displaystyle\int_{x}^{\infty} \dfrac{ dy'}{( 1+y')^{5/2}} \quad \color{#3D99F6}[y'=x+y]\\ & \quad\color{#20A900}[\text{By changing variables from } (x,y) \text{ to } (x,y')]\\ &=\frac{2}{5}\cdot\frac{2}{3}\int_{0}^{\infty} \dfrac{ dx}{( 1+x)^{3/2}}\\ &=\frac{2}{5}\cdot\frac{2}{3}\cdot\frac{2}{1}=\frac{8}{15}\end{aligned}$

So, $\boxed{a+b=8+15=23}$

Let $a = 1 + x + y + z$ . The integral becomes $\int_1^\infty da\ a^{-7/2} \iint_{0 \leq x,y}^{x+y \leq a-1} dx\ dy.$ The integration over $x$ and $y$ is equivalent to finding the area of the region defined by $x,y \geq 0$ and $x+y \leq a-1$ , which is a right triangle with legs $a-1$ , and therefore area $\tfrac12 (a-1)^2$ . Thus we continue $\int_1^\infty \frac12 (a-1)^2 a^{-7/2}\ da \\ = \int_1^\infty \left(\frac 12 a^{-3/2} - a^{-5/2} + \frac12 a^{-7/2}\right)\ da \\ = \left.\frac12\cdot \frac 1{-1/2} a^{-1/2} - \frac 1{-3/2} a^{-3/2} + \frac12\cdot \frac 1{-5/2} a^{-5/2}\right|_1^\infty \\ \frac12\cdot 2 - \frac 2 3 + \frac12\cdot \frac 2 5 = 1 - \frac 2 3 + \frac 1 5 = \frac 8{15}.$ Thus the answer is $8 + 15 = boxed{23}$ .