$\int_{0}^{\infty }\frac{1}{\sqrt{x}}e^{-x}dx$
- Tolaso J Kos
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$\int_{0}^{\infty }\frac{1}{\sqrt{x}}e^{-x}dx$
Evaluate the following integral:
$$\int_{0}^{\infty }\frac{1}{\sqrt{x}}e^{-x}\, {\rm d}x$$
$$\int_{0}^{\infty }\frac{1}{\sqrt{x}}e^{-x}\, {\rm d}x$$
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- Grigorios Kostakos
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Re: $\int_{0}^{\infty }\frac{1}{\sqrt{x}}e^{-x}dx$
\[\displaystyle \int_{0}^{\infty }\frac{1}{\sqrt{x}}e^{-x}dx=\int_{0}^{\infty }{x^{\frac{1}{2}-1}e^{-x}dx}=\Gamma\bigl({\tfrac{1}{2}}\bigr)\] From Euler's formula \(\Gamma(z)\,\Gamma(1-z)=\frac{\pi}{\sin(\pi\,z)}\) for \(z=\frac{1}{2}\) we have that \begin{align*}
\Gamma^2\bigl({\tfrac{1}{2}}\bigr)&=\Gamma\bigl({\tfrac{1}{2}}\bigr)\,\Gamma\bigl({1-\tfrac{1}{2}}\bigr)=\frac{\pi}{\sin\bigl({\frac{\pi}{2}}\bigr)}=\pi\quad\Rightarrow\\
\Gamma\bigl({\tfrac{1}{2}}\bigr)&=\sqrt{\pi}\,.
\end{align*} So \[\displaystyle \int_{0}^{\infty }\frac{1}{\sqrt{x}}e^{-x}dx=\sqrt{\pi}\,.\]
\Gamma^2\bigl({\tfrac{1}{2}}\bigr)&=\Gamma\bigl({\tfrac{1}{2}}\bigr)\,\Gamma\bigl({1-\tfrac{1}{2}}\bigr)=\frac{\pi}{\sin\bigl({\frac{\pi}{2}}\bigr)}=\pi\quad\Rightarrow\\
\Gamma\bigl({\tfrac{1}{2}}\bigr)&=\sqrt{\pi}\,.
\end{align*} So \[\displaystyle \int_{0}^{\infty }\frac{1}{\sqrt{x}}e^{-x}dx=\sqrt{\pi}\,.\]
Grigorios Kostakos
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Re: $\int_{0}^{\infty }\frac{1}{\sqrt{x}}e^{-x}dx$
2nd solution :
\(\displaystyle{\int_{0}^{\infty}\frac{e^{-x}}{\sqrt{x}}\,\rm{dx}=\int_{0}^{1}\frac{e^{-x}}{\sqrt{x}}\,\rm{dx}+\int_{1}^{\infty}\frac{e^{-x}}{\sqrt{x}}\,\rm{dx}}\) .
We define \(\displaystyle{F:\left(0,1\right]\longrightarrow \mathbb{R}}\) by \(\displaystyle{F(x)=\int_{x}^{1}\frac{e^{-t}}{\sqrt{t}}\,\rm{dt}}\) .
For each \(\displaystyle{x\in\left(0,1\right]}\) holds :
\(\displaystyle{\begin{aligned}\left|F(x)\right|&\leq \int_{x}^{1}\frac{e^{-t}}{\sqrt{t}}\,\rm{dt}\\&\leq \int_{x}^{1}\frac{e^{-x}}{\sqrt{t}}\,\rm{dt}\\&=\left[2\,e^{-x}\,\sqrt{t}\right]_{x}^{1}\\&=2\,e^{-x}\,\left(1-\sqrt{x}\right)\\&<2 \end{aligned}}\)
which means that the function \(\displaystyle{F}\) is bounded. So, \(\displaystyle{\int_{0}^{1}\frac{e^{-t}}{\sqrt{t}}\,\rm{dt}<\infty}\) .
Additionally, \(\displaystyle{\int_{1}^{\infty}\frac{e^{-t}}{\sqrt{t}}\,\rm{dt}<\infty}\) because
\(\displaystyle{\lim_{x\to +\infty}x^2\,\frac{e^{-x}}{\sqrt{x}}=\lim_{x\to +\infty}\frac{x^2}{e^{x}}\,\frac{1}{\sqrt{x}}=0}\) .
Thus, \(\displaystyle{\int_{0}^{\infty}\frac{e^{-x}}{\sqrt{x}}\,\rm{dx}=\int_{0}^{1}\frac{e^{-x}}{\sqrt{x}}\,\rm{dx}+\int_{1}^{\infty}\frac{e^{-x}}{\sqrt{x}}\,\rm{dx}<\infty}\) .
By applying the substitution \(\displaystyle{t=\sqrt{x}}\) , we have that
\(\displaystyle{t\in\left[0,+\infty\right)\,\,,x=t^2\Rightarrow \rm{dx}=2\,t\,\rm{dt}}\) , so
\(\displaystyle{\int_{0}^{\infty}\frac{e^{-x}}{\sqrt{x}}\,\rm{dx}=\int_{0}^{\infty}\frac{2\,t\,e^{-t^2}}{t}\,\rm{dt}=2\,\int_{0}^{\infty}e^{-t^2}\,\rm{dt}=2\,\frac{\sqrt{\pi}}{2}=\sqrt{\pi}}\) .
\(\displaystyle{\int_{0}^{\infty}\frac{e^{-x}}{\sqrt{x}}\,\rm{dx}=\int_{0}^{1}\frac{e^{-x}}{\sqrt{x}}\,\rm{dx}+\int_{1}^{\infty}\frac{e^{-x}}{\sqrt{x}}\,\rm{dx}}\) .
We define \(\displaystyle{F:\left(0,1\right]\longrightarrow \mathbb{R}}\) by \(\displaystyle{F(x)=\int_{x}^{1}\frac{e^{-t}}{\sqrt{t}}\,\rm{dt}}\) .
For each \(\displaystyle{x\in\left(0,1\right]}\) holds :
\(\displaystyle{\begin{aligned}\left|F(x)\right|&\leq \int_{x}^{1}\frac{e^{-t}}{\sqrt{t}}\,\rm{dt}\\&\leq \int_{x}^{1}\frac{e^{-x}}{\sqrt{t}}\,\rm{dt}\\&=\left[2\,e^{-x}\,\sqrt{t}\right]_{x}^{1}\\&=2\,e^{-x}\,\left(1-\sqrt{x}\right)\\&<2 \end{aligned}}\)
which means that the function \(\displaystyle{F}\) is bounded. So, \(\displaystyle{\int_{0}^{1}\frac{e^{-t}}{\sqrt{t}}\,\rm{dt}<\infty}\) .
Additionally, \(\displaystyle{\int_{1}^{\infty}\frac{e^{-t}}{\sqrt{t}}\,\rm{dt}<\infty}\) because
\(\displaystyle{\lim_{x\to +\infty}x^2\,\frac{e^{-x}}{\sqrt{x}}=\lim_{x\to +\infty}\frac{x^2}{e^{x}}\,\frac{1}{\sqrt{x}}=0}\) .
Thus, \(\displaystyle{\int_{0}^{\infty}\frac{e^{-x}}{\sqrt{x}}\,\rm{dx}=\int_{0}^{1}\frac{e^{-x}}{\sqrt{x}}\,\rm{dx}+\int_{1}^{\infty}\frac{e^{-x}}{\sqrt{x}}\,\rm{dx}<\infty}\) .
By applying the substitution \(\displaystyle{t=\sqrt{x}}\) , we have that
\(\displaystyle{t\in\left[0,+\infty\right)\,\,,x=t^2\Rightarrow \rm{dx}=2\,t\,\rm{dt}}\) , so
\(\displaystyle{\int_{0}^{\infty}\frac{e^{-x}}{\sqrt{x}}\,\rm{dx}=\int_{0}^{\infty}\frac{2\,t\,e^{-t^2}}{t}\,\rm{dt}=2\,\int_{0}^{\infty}e^{-t^2}\,\rm{dt}=2\,\frac{\sqrt{\pi}}{2}=\sqrt{\pi}}\) .
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