Jaqueline Kirby

Answered

2022-06-26

How do I show the following:

${\int}_{0}^{\alpha}dx{\int}_{0}^{\alpha}dy\frac{\mathrm{ln}|x-y|}{\sqrt{xy}}=4\alpha (\mathrm{ln}(\alpha )+2\mathrm{ln}(2)-3)$

where $\alpha >0$.

${\int}_{0}^{\alpha}dx{\int}_{0}^{\alpha}dy\frac{\mathrm{ln}|x-y|}{\sqrt{xy}}=4\alpha (\mathrm{ln}(\alpha )+2\mathrm{ln}(2)-3)$

where $\alpha >0$.

Answer & Explanation

Harold Cantrell

Expert

2022-06-27Added 21 answers

Proceed as follows

$\begin{array}{rl}{\int}_{0}^{a}{\int}_{0}^{a}\frac{\mathrm{ln}|x-y|}{\sqrt{xy}}dy\phantom{\rule{mediummathspace}{0ex}}dx& =2{\int}_{0}^{a}{\int}_{0}^{x}\frac{\mathrm{ln}(x-y)}{\sqrt{xy}}dy\phantom{\rule{mediummathspace}{0ex}}dx\\ & \stackrel{y=x{t}^{2}}{=}4{\int}_{0}^{a}{\int}_{0}^{1}(\mathrm{ln}x+\mathrm{ln}(1-{t}^{2}))dt\phantom{\rule{mediummathspace}{0ex}}dx\\ & =4{\int}_{0}^{a}(\mathrm{ln}x+2\mathrm{ln}2-2)\phantom{\rule{mediummathspace}{0ex}}dx\\ & =4a\phantom{\rule{mediummathspace}{0ex}}(\mathrm{ln}a+2\mathrm{ln}2-3)\end{array}$

$\begin{array}{rl}{\int}_{0}^{a}{\int}_{0}^{a}\frac{\mathrm{ln}|x-y|}{\sqrt{xy}}dy\phantom{\rule{mediummathspace}{0ex}}dx& =2{\int}_{0}^{a}{\int}_{0}^{x}\frac{\mathrm{ln}(x-y)}{\sqrt{xy}}dy\phantom{\rule{mediummathspace}{0ex}}dx\\ & \stackrel{y=x{t}^{2}}{=}4{\int}_{0}^{a}{\int}_{0}^{1}(\mathrm{ln}x+\mathrm{ln}(1-{t}^{2}))dt\phantom{\rule{mediummathspace}{0ex}}dx\\ & =4{\int}_{0}^{a}(\mathrm{ln}x+2\mathrm{ln}2-2)\phantom{\rule{mediummathspace}{0ex}}dx\\ & =4a\phantom{\rule{mediummathspace}{0ex}}(\mathrm{ln}a+2\mathrm{ln}2-3)\end{array}$

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