一道与黄金分割数有关的广义积分
[i=s] 本帖最后由 青青子衿 于 2019-5-14 18:26 编辑 [/i]设$\,\varphi = \dfrac{\sqrt{5}+1}{2}\,$，求证:
\[\color{black}{\int_0^\infty\frac{1}{\left(t+\sqrt{1+t^2}\,\right)^\varphi}\mathrm{d}t = 1} \]
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\[ \displaystyle\color{black}{I(\alpha,\beta) = \int_0^\alpha\left(x+\sqrt{x^2+1}\right)^\beta \, \mathrm{d}x} \]
\begin{align*}
u&= \beta\operatorname{arsinh}\left(x\right)\\
&= \beta\ln\left(x+\sqrt{x^2+1}\,\right)\\
&= \ln\left[\left(x+\sqrt{x^2+1}\,\right)^\beta\right]\\
e^u&= \left(x+\sqrt{x^2+1}\,\right)^\beta\\
x&=\sinh\left(\frac{u}{\beta}\right)\\
\mathrm{d}x&= \dfrac{1}{\beta}\cosh\left(\frac{u}{\beta}\right)\mathrm{d}u
\end{align*}
\begin{align*}
I(\alpha,\beta)
&= \int_0^{\beta\operatorname{arsinh}\left(\alpha\right)} e^u \cdot \frac{1}{\beta} \cosh \left(\frac{u}{\beta} \right) \, \mathrm{d}u \\
&= \int_0^{\beta\ln\left(\alpha+\sqrt{\alpha^2+1}\,\right)} e^u \cdot \frac{1}{\beta} \cosh \left(\frac{u}{\beta} \right) \, \mathrm{d}u
\,\,\overset{\gamma\,\,\triangleq\,\beta\operatorname{arsinh}\left(\alpha\right)}{\overline{\overline{\hspace{2.3cm}}}}\,
\int_0^\gamma e^u \cdot \frac{1}{\beta} \cosh \left(\frac{u}{\beta} \right) \, \mathrm{d}u \\
&= \int_0^\gamma e^u\cdot\left(\dfrac{e^{u/\beta}+e^{-u/\beta}}{2\beta}\right) \, \mathrm{d}u
= \frac{1}{2\beta} \int_0^\gamma \left[e^{(\beta+1)u/\beta} + e^{(\beta-1)u/\beta} \right]\mathrm{d}u \\
&= \frac{1}{2\beta} \left[\frac{\beta}{\beta+1}e^{(\beta+1)u/\beta} + \frac{\beta}{\beta-1}e^{(\beta-1)u/\beta} \right]_0^\gamma \\
&= \frac{1}{2\beta} \left[\frac{\beta}{\beta+1}e^{(\beta+1)\gamma/\beta} + \frac{\beta}{\beta-1} e^{(\beta-1)\gamma/\beta} - \left(\frac{\beta}{\beta+1}+\frac{\beta}{\beta-1} \right) \right] \\
&= \frac{1}{2\beta} \left[\frac{\beta}{\beta+1}e^{(\beta+1)\ln\left(\alpha+\sqrt{\alpha^2+1}\,\right)} + \frac{\beta}{\beta-1}e^{(\beta-1)\ln\left(\alpha+\sqrt{\alpha^2+1}\,\right)} - \frac{2\beta^2}{(\beta+1)(\beta-1)} \right] \\
&= \frac{\left(\alpha+\sqrt{\alpha^2+1}\right)^{\beta+1}}{2(\beta+1)} + \dfrac{\left(\alpha+\sqrt{\alpha^2+1}\right)^{\beta-1}}{2(\beta-1)} - \frac{\beta}{(\beta+1)(\beta-1)} \end{align*}
[url]https://www.quora.com/How-can-I-show-displaystyle-lim_-n-to-infty-n-int_0-1-n-x-sqrt-x-2-1-n-mathrm-d-x-e-1[/url] [b]回复 [url=http://kuing.orzweb.net/redirect.php?goto=findpost&pid=31293&ptid=6101]1#[/url] [i]青青子衿[/i] [/b]
\[\color{black}{\int_0^\infty\frac{x^{p-1}}{\left(t+\sqrt{1+t^2}\,\right)^q}\mathrm{d}t =\dfrac{q\,\Gamma(p)\,\Gamma\left(\frac{q-p}{2}\right)}{2^{p+1}\,\Gamma\left(\frac{p+q}{2}+1\right)}} \]
Berndt, B. C.; Ramanujan's Notebooks, Part I,
Springer-Verlag, New York, 1985.
P305
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