Compute $$\int^{\infty}_{-\infty}{\int^{\infty}_{0}{xe^{-x^2e^y}}}dydx$$ Since x is just a constant in the inner integral I took it out and was only left with $\int^{\infty}_{0}{}e^{-x^2e^y}dy.$ This is where I'm stuck. I cannot figure out this integral. Could somebody help me with this if I'm in the right track, or am I completely wrong? I feel like this lacks an easy solution therefore the result might be complex.
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There's indeed a very quick solution. Hint : the final result is $0$. – Zag Jan 08 '23 at 15:05
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Please make the title more descriptive. – mr_e_man Jan 08 '23 at 23:16
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You've gotten a lot of good answers, you should probably accept one of them :) – Lorago Jan 10 '23 at 17:42
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First notice that the exchange of the order of integration is allowed by continuity, then $$\int_{-\infty}^{+\infty}\int_{0}^{+\infty}xe^{-x^{2}e^{y}}\, dydx=\int_{0}^{+\infty}\int_{-\infty}^{+\infty}xe^{-x^{2}e^{y}}\, dxdy.$$ But if $\phi(x):=xe^{-x^{2}e^{y}}$, then $\phi(-x)=-\phi(x)$ so we have a odd function over all the interval $]-\infty,+\infty[$. Can you continue from here?
A. P.
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I can't, thats completely new to me. Also why phi function is $x^2e^{-x^{2}e^{y}}$. Shouldn't it be just $x$. – Techlover Jan 08 '23 at 15:49
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That was a type. I fixed it, just right now. Since $\phi$ is odd function, then $\int_{-a}^{a}\phi(x), dx=0$. Then, your double integral is zero. – A. P. Jan 08 '23 at 15:50
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It it is a theorem: If $\phi$ is odd function over $[-a,a]$, then $\int_{-a}^{a}\phi(x), dx=0$. Think about it geometrically: since $\phi$ is odd, then it is symmetric with the origin, then the integral on the interval $[-a,0]$ is the negative of the integral on $[0,a]$, therefore the integral on $[-a,a]$ is zero. – A. P. Jan 08 '23 at 15:57
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If we set
$$f(x)=\int_0^\infty xe^{-x^2e^y}~\mathrm{d}y,$$
then
$$f(-x)=\int_0^\infty -xe^{-(-x)^2e^y}~\mathrm{d}y=-\int_0^\infty xe^{-x^2e^y}~\mathrm{d}y=-f(x),$$
i.e. $f$ is an odd function. Now showing that your integral
$$I=\int_{-\infty}^\infty\int_0^\infty xe^{-x^2e^y}~\mathrm{d}y~\mathrm{d}x=\int_{-\infty}^\infty f(x)~\mathrm{d}x$$
converges is very straightforward, and so we can compute it as its principal value. Thus, as the integral of an odd function over a symmetric interval is $0$, we finally get that
$$I=\lim_{c\to\infty}\underbrace{\int_{-c}^c f(x)~\mathrm{d}x}_{\text{$0$ as $f$ is odd}}=0.$$
Lorago
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$$ \begin{aligned} \int_0^{+\infty}e^{-x^2e^y}dy &= \int_0^{+\infty}\frac{e^{-x^2e^y}}{x^2e^y}x^2e^ydy = \begin{vmatrix} u = x^2e^y & \Rightarrow & du = x^2e^ydy \\ y = 0 &\Leftrightarrow &u = x^2 \\ y = +\infty &\Leftrightarrow &u = +\infty \end{vmatrix} = \\ &= \int_{x^2}^{+\infty}\frac{e^{-u}}{u}du = \varphi\left(x^2\right), \end{aligned} $$
where $\varphi(\cdot)$ is some function.
Then,
$$ \begin{aligned} \int_{-\infty}^{+\infty}\int_0^{+\infty}xe^{-x^2e^y}dydx &= \int_{-\infty}^{+\infty}x\varphi\left(x^2\right)dx = 0, \end{aligned} $$
since it is an integral of an odd function $\left(x\varphi\left(x^2\right)\right)$ over the symmetric interval $(-\infty, +\infty)$.
Eugene
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HINT: The inner integral over y is the incomplete Euler $\Gamma$ function
$$\int_0^{\infty } \exp \left(-x^2 \exp (y)\right) \, dy=\fbox{$\Gamma \left(0,x^2\right)\text{ if }\Re\left(x^2\right)>0$}$$
The outer integral over x results to
$$F(x)=\int x \Gamma \left(0,x^2\right) \, dx=\frac{1}{2} x^2 \Gamma \left(0,x^2\right)-\frac{e^{-x^2}}{2}$$
Now we use the integration limits
$$\underset{x\to \infty }{\text{lim}}F(x)-\underset{x\to -\infty }{\text{lim}}F(x)=0$$
gpmath
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