My Math Forum Calculus: Integration

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 February 17th, 2018, 09:58 PM #1 Newbie   Joined: Feb 2018 From: India Posts: 5 Thanks: 0 Math Focus: calculas, linear algebra, geometry, complex analysis,polynomials Calculus: Integration ${\displaystyle\int_0^1}\! \sqrt{x\ln\small\left(\frac{1}{\large x}\right)}dx$ Last edited by greg1313; February 17th, 2018 at 10:56 PM.
 February 17th, 2018, 10:57 PM #2 Global Moderator   Joined: Dec 2006 Posts: 18,719 Thanks: 1536 According to W|A, it's $\displaystyle \frac13\sqrt{\frac{2\pi}{3}}$.
 February 18th, 2018, 07:08 AM #3 Newbie   Joined: Feb 2018 From: India Posts: 5 Thanks: 0 Math Focus: calculas, linear algebra, geometry, complex analysis,polynomials elaborate solution
 February 18th, 2018, 08:21 AM #4 Newbie   Joined: Feb 2018 From: India Posts: 5 Thanks: 0 Math Focus: calculas, linear algebra, geometry, complex analysis,polynomials also, if square root is replaced by cube root what will be the solution?
 February 18th, 2018, 09:42 AM #5 Global Moderator   Joined: Dec 2006 Posts: 18,719 Thanks: 1536 Have you tried using a substitution? Thanks from topsquark
February 18th, 2018, 09:43 AM   #6
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Quote:
 Originally Posted by skipjack Have you tried using a substitution?
or anything at all?

 February 18th, 2018, 06:30 PM #7 Global Moderator   Joined: Dec 2006 Posts: 18,719 Thanks: 1536 Using $x = e^{-u^2}$, where $u\geqslant0$, so that $dx = -2ue^{-u^2}du$, one gets $\displaystyle \int_\infty^0\! -2u^2e^{-3u^2/2}du = -(2/3)\!\int_0^\infty\! -3u^2e^{-3u^2/2}du$. Using integration by parts, with $v = e^{-(3/2)u^2}$, so that $dv = -3ue^{-(3/2)u^2}du$, one gets $\displaystyle -(2/3)\left[ue^{-(3/2)u^2}\right]_0^\infty + (2/3)\!\int_0^\infty\! e^{-(3/2)u^2}du = (2/3)\!\int_0^\infty\! e^{-(3/2)u^2}du$. If $u = \sqrt{\frac23}t$, so that $du = \sqrt{\frac23}dt$, one gets $\frac23\sqrt{\frac23}$ $\displaystyle \!\int_0^\infty\! e^{-t^2}dt$. It's well known that $\displaystyle \!\int_0^\infty\! e^{-t^2}dt = \small\frac{\sqrt\pi}{2}$. Thanks from Country Boy

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