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I asked Wolfram Alpha what $$ \int_0^\infty \exp\bigl(-\sqrt x\bigr)\, dx $$ and at the first reshape it says that it's substituting $\sqrt x$ to $x$, $dx$ to $1/2\sqrt x \,dx$ and that the boundaries stay the same. But then it says, the asked integral is equal to $$ 2\int_0^\infty \exp(-x)x\, dx $$ I don't understand why it's $$\int_0^\infty 2x\exp(-x)\, dx$$ and not $$ \int_0^\infty \frac 1{2\sqrt x}\exp(-x) \, dx.$$ Can somebody say why?

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  • $\begingroup$ WA tells me that the answer to the original integral is $\Gamma(5/3)$... $\endgroup$ – complexmanifold Jan 3 '17 at 13:06
  • $\begingroup$ @Bacon: double check your result. $\endgroup$ – Yves Daoust Jan 3 '17 at 13:14
  • $\begingroup$ @YvesDaoust Whoops! It is equal to $2$. $\endgroup$ – complexmanifold Jan 3 '17 at 13:22
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You messed up the variables… don't use $x$ if you substitute $\sqrt{x}$ then this won't happen.

So: $y = \sqrt{x}$ then $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$ and so $dy = \frac{1}{2\sqrt{x}} dx$ or equivalent $2ydy = dx$ and hence

$$\int_0^\infty \exp\bigl(-\sqrt x\bigr)\, dx = \int_0^\infty 2y \exp\bigl(-y\bigr)\, dy$$

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  • $\begingroup$ O.K., I see, thank you for your answer! Also int, 0 to inf, e^-(x^n)dx = (1/n)!, so with n=(1/2) we get 2! = 2 $\endgroup$ – cb76 Jan 4 '17 at 15:04
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Let $x = u^2$; then $dx = 2u du$, and as $x$ goes from $0$ to $\infty$, so does $u$. Your integral becomes $$ \int_0^\infty e^{-\sqrt{u^2}} 2u~du = \int_0^\infty 2e^{-u} u~du= \int_0^\infty 2e^{-x} x~dx. $$

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To avoid confusion, use different letters.

With $z=\sqrt x$, we have $x=z^2$ and $dx=2z\,dz$. Then

$$\int e^{-\sqrt x}dx=2\int e^{-z}z\,dz.$$


The integral is easy, by parts,

$$\int e^{-z}z\,dz=-e^{-z}z+\int e^{-z}dz=-e^{-z}z-e^{-z},$$ giving $1$ between $0$ and $\infty$.

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