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Since $e^{-x^{2}/2}$ is an even function. $\int_{0}^{\infty}e^{-x^{2}/2} dx=\sqrt{\pi/2}.$

Similarly, as $|x|^{1/2}e^{|x|}$ an even function. $\int_{-\infty}^{\infty}|x|^{1/2}e^{|x|} dx=2\int_{0}^{\infty}\sqrt{x}e^{x} dx$

$\mathbf{My \ attempt}$

Let $x=t^{2}/2$ $\Rightarrow dx=t\ dt$

Therefore, $2\int_{0}^{\infty}\sqrt{x}e^{x} dx=\sqrt{2}\int_{0}^{\infty}t^{2}e^{t^{2}/2}dt$.

Which is where I get stuck. I believe there's a misprint in the question. The power of the $|x|$ should be $-\frac{1}{2}$ and not $\frac{1}{2}$.

Because the I'll get $2\int_{0}^{\infty}\sqrt{x}e^{x} dx=\sqrt{2}\int_{0}^{\infty}e^{t^{2}/2}dt=\sqrt{2} \cdot \sqrt{\pi/2}=\sqrt{\pi}$.

But I may be wrong. Please point me in the right direction. Thanks in advance.

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    There are at least a couple things wrong here, possibly just typos. The exponent in $e^{x^2/2}$ should be $-x^2/2$ because if it's $x^2/2$ then the integral diverges. Similarly, the $e^{|x|}$ needs to be $e^{-|x|}$. As long as you have $e^{-|x|}$ then $|x|^{1/2}$ is fine. – 2017-02-28
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    Unfortunately, they aren't typos. As in, that's how the question has been given. I think the question has a lot of issues. – 2017-02-28
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    If really this is what you were given, you can stop at "Given the value of $\int_{-\infty}^{\infty}e^{x^{2}/2} dx=\sqrt{2\pi}$", which is quite absurd. Given this, I am the Pope and $1=0$... – 2017-02-28
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    @junkquill12, what Did said. Maybe there aren't typos in what you typed, but if that's the question you were given, then the question has typos. – 2017-02-28
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    The actual title makes no sense at all: $$\int_{0}^{R}\sqrt{x}\,e^x\,dx$$ is clearly diverging as $R\to +\infty$. – 2017-02-28
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    Yes, I agree. I failed to see that when I first typed in the question. I think the question itself is bogus then. Thanks for all the responses guys! – 2017-02-28
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    Do you mean $\int_{-\infty}^{\infty}|x|^{\frac12}e^{-|x|}dx$ instead of $\int_{-\infty}^{\infty}|x|^{\frac12}e^{|x|}dx$? – 2017-02-28

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