Triple Trouble!

Calculus Level 4

$\displaystyle\int_0^{\infty}\displaystyle\int_0^{\infty}\displaystyle\int_0^{\infty} \dfrac{\ dx \ dy \ dz}{\sqrt{(1+x+y+z)^7}}$

If the above integral can be expressed as $\dfrac{S}{G}$ . Find $\overline{SG}$

$\implies$ S,G are coprime integers

$\implies$ If the answer is 17/15 the you should input 1715.

The answer is 815.

1 solution

Aditya Kumar
Jun 23, 2015

$I=\int _{ 0 }^{ \infty }{ \int _{ 0 }^{ \infty }{ \int _{ 0 }^{ \infty }{ \frac { dx.dy.dz }{ \sqrt { { (1+x+y+z) }^{ 7 } } } } } } \\ =\int _{ 0 }^{ \infty }{ \int _{ 0 }^{ \infty }{ \left( \int _{ 0 }^{ \infty }{ \frac { dx }{ \sqrt { { (1+x+y+z) }^{ 7 } } } } \right) dy.dz } } \\ Let\quad { I }_{ 1 }=\int _{ 0 }^{ \infty }{ \frac { dx }{ \sqrt { { (1+x+y+z) }^{ 7 } } } } \\ Let\quad 1+x+y+z=t\quad \quad \\ \therefore \quad { I }_{ 1 }=\int _{ 0 }^{ \infty }{ \frac { dt }{ \sqrt { { (t) }^{ 7 } } } } \\ \Rightarrow { I }_{ 1 }=\frac { -2 }{ 5 } \frac { 1 }{ \sqrt { { (1+y+z) }^{ 5 } } } \\ Similarly\quad solve\quad the\quad other\quad part\quad of\quad the\quad integral.\\ \therefore I=\frac { 8 }{ 15 }$

@Parth Lohomi can u suggest any simpler method, if u have?

Aditya Kumar - 5 years, 11 months ago

@Aditya Kumar , can you explain your solution? I don't understand your solution at all.