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Why the definite integral of: $$-ze^{-\frac{z^2}{2}}$$ Evaluated from 0 to $\infty$ equal 0 ????

I plot the graph of the function and see some area but when integrate to find its area it turns out to be 0 ??? why the Algebraic result contradict with geometric result? isn't the area suppose to be negative? ex. we can check the value when $ z \in (0,2] $ of this function and it is all negative. when z > 3 it seems to approach 0 so the area should be negative isn't? why it isn't?

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    is $z$ here a complex number? – 2017-02-22
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    What algebraic result? – 2017-02-22
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    This integral is not zero. – 2017-02-22
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    You should include your calculation that the integral is $0$. That's where the problem lies. – 2017-02-22
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    The integral results in -1 – 2017-02-22
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    ok i will check with it again thank you very much ^^" – 2017-02-22

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I'm not sure what you mean, but if we let $u=\frac{z^2}2$, we get the "algebraic" result:

$$\int_0^\infty-ze^{-\frac{z^2}2}\ dz=\int_0^\infty-e^{-u}\ du=e^{-u}\bigg|_{u=0}^\infty=e^{-\infty}-e^0=0-1=-1$$

Clearly, this is negative...