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Can someone please show how to calculate this with integration by parts $(\int udv = uv - \int vdu)$? I found an example in the book is not clear and confusing.

$$\frac{2}{\sqrt{2\pi}} \int^\infty_0 z^2e^{\frac{-z^2}{2}}dz$$

They start with u = z, dv = $ze^{\frac{-z^2}{2}} dz$, v = $ -e^{\frac{-z^2}{2}}$, du = dz

then:

$$\int^\infty_0 udv = -ze^{\frac{-z^2}{2}} - \int^\infty_0 -e^{\frac{-z^2}{2}} dz = -ze^{\frac{-z^2}{2}} + \int^\infty_0 e^{\frac{-z^2}{2}} = -ze^{\frac{-z^2}{2}} + \frac{\sqrt{2\pi}}{2}$$

I have no problem with the most right part of integration $\int^\infty_0 e^{\frac{-z^2}{2}}$ i know it is equal to$\frac{\sqrt{2\pi}}{2}$ but what about the $-ze^{\frac{-z^2}{2}}$ how to deal with it?

how can I continue from here?

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    Don't worry, it is normal that you are wondering how to continue, as the undefined integral has no closed-form, unless you heard of the "error function" or the Gaussian integral. Depending on how advanced your courses are, the question is fair or unfair. – 2017-02-22
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    You should have said that in the first place. – 2017-02-22
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    ok i will update the question. Thank you – 2017-02-22
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    Integration by parts formula: $$\int_a^b u\mathrm dv=uv\big|_a^b-\int_a^b v\mathrm du$$ ... – 2017-02-22
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    Another possible way to evaluate it: if we set $f(a) = \frac{2}{\sqrt{2\pi}}\int_0^\infty e^{- az^2}$ then your integral is $-\left.\frac{df(a)}{da}\right|_{a = 1/2}$ and $f(a)$ is a [Gaussian integral](https://en.wikipedia.org/wiki/Gaussian_integral). – 2017-02-22

2 Answers 2

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$$\left.-ze^{-z^2/2}\right|_0^\infty=0-0.$$

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    why I have to evaluate it from 0 to $\infty$ ? it just a function isn't it? I'm not familiar with integration by part. That part is not in the integration???? – 2017-02-22
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    @user3270418: this is a definite integral, handle it like a definite integral. – 2017-02-22
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With $\dfrac{z^2}{2}=u$ and $$\frac{2}{\sqrt{2\pi}} \int^\infty_0 z^2e^{\frac{-z^2}{2}}dz=\frac{2}{\sqrt{\pi}} \int^\infty_0 u^\frac12e^{-u}du=\frac{2}{\sqrt{\pi}} \Gamma(\frac32)=\frac{2}{\sqrt{\pi}} \frac{\sqrt{\pi}}{2}=1$$

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    There is virtually no chance the OP knows better the Gamma function – 2017-02-22