I have some problem with the result of this integral:

$$\int_{0}^{+\infty} e^{-\sqrt{x}}dx$$

The result should be 2 but I get -2 but I do not see the error, can someone please show me where is it?

$$\lim_{k\rightarrow \infty} \int_{0}^{k} e^{-\sqrt{x}} dx$$


$$-\sqrt{x} = t$$


$$ dx = 2t $$

Calculate the indefinite integral:

$$ \int e^{-\sqrt{x}} dx = 2\int te^t dt$$

Solve by parts:

$$2\int te^t dt = 2te^t - 2e^t$$

Get back in x:

$$2te^t - 2e^t = -2\sqrt{x}e^{-\sqrt{x}} - 2e^{-\sqrt{x}}$$


$$ \lim_{k\rightarrow \infty} \big[ -2\sqrt{x}e^{-\sqrt{x}} - 2e^{-\sqrt{x}} \big]_{0}^{k} = -2$$

  • $\begingroup$ In general, $~\displaystyle\int_0^\infty e^{\large-\sqrt[n]x}~dx~=~n!$ $\endgroup$ – Lucian Jun 24 '15 at 10:06

Note that when substituting the limits the only non-vanishing term comes from the lower limit ($x=0$) and hence changes sign.

  • $\begingroup$ Yeah, thanks...just a stupid error! $\endgroup$ – Christian Giupponi Jun 24 '15 at 8:55

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