Question

$\displaystyle\tt \int_{0}^{ \infty } {xe}^{ - \sqrt{x} } \: ln(x) \: dx$​

Answer 1

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Answer 2

To solve the given integral, we will use integration by parts with u = ln x and dv = x $e^{-\sqrt{x}} dx$. Then, du/dx = 1/x and v = $-2 e^{-\sqrt{x}} (\sqrt{x} + 1)$.

Using the formula for integration by parts, we have:

$\begin{align}\int_0^\infty x e^{-\sqrt{x}} \ln x \, dx &= \left[-2xe^{-\sqrt{x}} (\sqrt{x} + 1) \ln x \right]_0^\infty + \int_0^\infty 2e^{-\sqrt{x}} (\sqrt{x} + 1) \, dx \\&= \int_0^\infty 2e^{-\sqrt{x}} \sqrt{x} \, dx + \int_0^\infty 2e^{-\sqrt{x}} \, dx \\\end{align}$

We can evaluate the first integral using the substitution u = √x, which gives:

$\begin{align}\int_0^\infty 2e^{-\sqrt{x}} \sqrt{x} \, dx &= 4 \int_0^\infty e^{-u} u^2 \, du \\&= 4 \Gamma(3) \\&= 8 \\\end{align}$

where Γ is the gamma function.

The second integral can be evaluated directly as:

$\begin{align}\int_0^\infty 2e^{-\sqrt{x}} \, dx &=\left[-4e^{-\sqrt{x}} \right]_0^\infty \\&= 4 \\ \end{align}$

Therefore, the value of the integral is:

$\begin{align}\int_0^\infty x e^{-\sqrt{x}} \ln x \, dx &= \int_0^\infty 2e^{-\sqrt{x}} \sqrt{x} \, dx + \int_0^\infty 2e^{-\sqrt{x}} \, dx \\&= 8 + 4 \\&= 12 \\\end{align}$

Thus, the value of the given integral is 12.