# Evaluate some kind of gamma function

#### ozgunozgur

##### New member
My question and solution that I've tried out are in attachment. Is it true my steps?

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#### Country Boy

##### Well-known member
MHB Math Helper
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.