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Evaluate some kind of gamma function

ozgunozgur

New member
Jun 1, 2020
27
My question and solution that I've tried out are in attachment. Is it true my steps?
 

Attachments

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
426
I am puzzled that you post the problem as $\int_0^1\int_{\sqrt{x}}^1e^{y^y} dxdy$, which makes no sense to me but do it as $\int_0^1\int_{\sqrt{x}}^1e^{y^3} dydx$ which does make sense and is far easier! Which is it?

The first integral makes no sense because the integral with respect to x has a function of x as the lower bound so that, even after the first integral, you will have a function of both x and y and after integrating with respect to y you will still have a function of x instead of a number.

And the second integral is far easier because $e^{y^y}$ is a horrendous function to integrate while $e^{y^3}$ is much easier!

To integrate $\int_0^1\int_{\sqrt{x}}^1e^{y^3} dydx$, I would first change the order of integration. The integral, taking x from 0 to 1 and, for each x, y from $\sqrt{x}$ to 1, is the portion of the square $0\le x\le 1$, $0\le y\le 1$, above the graph of $y= \sqrt{x}$. That is also the portion to the right of $x= y^2$ so the integral is $\int_0^1\int_0^{y^2} e^{y^3}dxdy$. There is no "x" in the integrand so the first integral just results in $\int_0^1 e^{y^3}y^2e^{y^3}dy$ which is easy.