# wrong result for $\int_{0}^{+\infty} e^{-\sqrt{x}}dx$

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?

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

let

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

so:

$$dx = 2t$$

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:

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

So:

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

• In general, $~\displaystyle\int_0^\infty e^{\large-\sqrt[n]x}~dx~=~n!$ – 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.