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integral related to arc sine
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(Example)
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We want to evaluate the integral
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(1) |
Therefore we put an extra variable $t$ to the integrand and thus get the function $$I(t) \;:=\; \int_1^\infty\!\left(\arcsin\frac{t}{x}-\frac{t}{x}\right)dx,$$ and in order to obtain a simpler integral, we differentiate it under the integral sign, then integrate:
The gotten expression implies, since $I(0) = \int_1^\infty(\arcsin{0}-0)dx = 0$ , that $$I(t) \;=\; \int_0^t[\ln{2}-\ln(1+\sqrt{1\!-\!t^2})]\,dt \;=\; t\ln{2}-\int_0^t\ln(1+\sqrt{1\!-\!t^2})\,dt,$$ and consequently
Here, the substitution $t = \sin{u}$ helps, yielding $$I(1) \;=\; \ln{2}-\int_0^{\frac{\pi}{2}}\!(1-\cos{u})\,du \;=\; \ln{2}-\frac{\pi}{2}+1.$$ Accordingly, we have the result $$\int_1^\infty\!\left(\arcsin\frac{1}{x}-\frac{1}{x}\right)dx \;=\; 1+\ln{2}-\frac{\pi}{2}.$$
For the convergence, see the French version of this article.
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"integral related to arc sine" is owned by pahio.
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Cross-references: implies, expression, integrate, function, integrand, variable, integral
There are 2 references to this entry.
This is version 5 of integral related to arc sine, born on 2009-01-19, modified 2009-01-20.
Object id is 11526, canonical name is IntegralRelatedToArcSine.
Accessed 547 times total.
Classification:
| AMS MSC: | 26A09 (Real functions :: Functions of one variable :: Elementary functions) |
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Pending Errata and Addenda
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![$\displaystyle \;=\; \operatornamewithlimits{\Big/}_{\!\!\!x=1}^{\,\quad\infty}\!\left[\ln\left(\frac{x}{t}+\sqrt{\frac{x^2}{t^2}\!-\!1}\right)-\ln{x}\right]$](http://images.planetmath.org:8080/cache/objects/11526/js/img5.png "$\displaystyle \;=\; \operatornamewithlimits{\Big/}_{\!\!\!x=1}^{\,\quad\infty}\!\left[\ln\left(\frac{x}{t}+\sqrt{\frac{x^2}{t^2}\!-\!1}\right)-\ln{x}\right]$")
