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Calculus Level pending

Find $\displaystyle \int_{x=0}^{\infty} dx \int_{y=0}^{1} dy \int_{z=0}^{x} dz x e^{\left (-2\dfrac{x}{y}-3xy-5yz \right )} =K$ .

Enter your answer as $\displaystyle \dfrac{1}{K}$

Bonus: Try generalise

$\displaystyle \int_{x=0}^{\infty} dx \int_{y=0}^{1} dy \int_{z=0}^{x} dz x e^{\left (-A\dfrac{x}{y}-Bxy-Cyz \right )}$

For positive A,B and C.

1 solution

คลุง แจ็ค
Apr 20, 2017

Here is another way of doing it!!

Once we have done the z integral, then we are left with integrals of the form

$\displaystyle \int_{x=0}^{\infty} \int_{y=0}^{1} \dfrac{-x}{cy} \exp{(-(A\dfrac{x}{y}+Bxy))} dydx$

Knowing that the order of integration can be changed, thus

$\displaystyle \int_{0}^{\infty} x \exp{(-Ax)} dx=\dfrac{1}{A^{2}}$ So, we are left with

$\displaystyle \int_{0}^{1} \dfrac{y}{(By^{2}+A)^{2}} dy=\dfrac{1}{2A(A+B)}$ So,

$\displaystyle \int_{x=0}^{\infty} \int_{y=0}^{1} \int_{z=0}^{x} x \exp{(-(A\dfrac{x}{y}+Bxy+Cyz))} dzdydx =\boxed{ \dfrac{1}{2A(A+B)(A+B+C)}}$