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  #1  
Old December 2nd, 2008, 07:24 AM
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Default Integrate Sqrt[x^2-a]?
I believe the answer is 0.5(x(x^2-a)^0.5)-0.5aln(x+(x^2-a)^0.5)

does anyone know how to get this/derive this i.e. not take from tables !!!!

Thanks
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  #2  
Old December 2nd, 2008, 08:05 AM
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Originally Posted by coverband View Post
I believe the answer is 0.5(x(x^2-a)^0.5)-0.5aln(x+(x^2-a)^0.5)

does anyone know how to get this/derive this i.e. not take from tables !!!!

Thanks
that is correct.

to get the derivative of the answer, you need to use the product rule and the chain rule. are you familiar with these.

moreover, note that \frac d{dx} \ln u = \frac {u'}u (by the chain rule)

where u is some function of x
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  #3  
Old December 2nd, 2008, 08:39 AM
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Originally Posted by Jhevon View Post
that is correct.

to get the derivative of the answer, you need to use the product rule and the chain rule. are you familiar with these.

moreover, note that \frac d{dx} \ln u = \frac {u'}u (by the chain rule)

where u is some function of x
I could be wrong, but I think s/he was asking how the result was reached.
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Old December 4th, 2008, 01:53 PM
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try letting x=\sqrt{a}\cosh(\theta)
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\int_0^z\ln\Gamma(x+1)dx=\frac{z}{2}\ln(2\pi)-\frac{z(z+1)}{2}+z\ln\Gamma(z+1)-\ln\left\{(2\pi)^{\frac{z}{2}}\exp\left(\frac{-z(z+1)}{2}-\frac{\gamma z^2}{2}\right)\prod_{k=1}^{\infty}\left\{\left(1+\frac{z}{k}\right)^k\exp\left(-z+\frac{z^2}{2k}\right)\right\}\right\}
Last edited by mr fantastic; December 4th, 2008 at 08:21 PM.
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Old December 4th, 2008, 01:56 PM
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mathstud what if a is negative - then we can't consider sqrt(a) ?
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Old December 4th, 2008, 01:57 PM
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Originally Posted by coverband View Post
mathstud what if a is negative - then we can't consider sqrt(a) ?
Ok then make it \sqrt{x^2+(-a)} and make the sub x=\sqrt{-a}\tan(\theta) or x=\sqrt{-a}\sinh(\theta)
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\int_0^z\ln\Gamma(x+1)dx=\frac{z}{2}\ln(2\pi)-\frac{z(z+1)}{2}+z\ln\Gamma(z+1)-\ln\left\{(2\pi)^{\frac{z}{2}}\exp\left(\frac{-z(z+1)}{2}-\frac{\gamma z^2}{2}\right)\prod_{k=1}^{\infty}\left\{\left(1+\frac{z}{k}\right)^k\exp\left(-z+\frac{z^2}{2k}\right)\right\}\right\}
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Old December 4th, 2008, 01:59 PM
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but we want to integrate Sqrt[x^2-a] and i forsee difficulty for if a<0 we can not use sqrt(a) in our substitutions
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Old December 4th, 2008, 02:00 PM
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but we want to integrate Sqrt[x^2-a] and i forsee difficulty for if a<0 we can not use sqrt(a) in our substitutions
I edited my post
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\int_0^z\ln\Gamma(x+1)dx=\frac{z}{2}\ln(2\pi)-\frac{z(z+1)}{2}+z\ln\Gamma(z+1)-\ln\left\{(2\pi)^{\frac{z}{2}}\exp\left(\frac{-z(z+1)}{2}-\frac{\gamma z^2}{2}\right)\prod_{k=1}^{\infty}\left\{\left(1+\frac{z}{k}\right)^k\exp\left(-z+\frac{z^2}{2k}\right)\right\}\right\}
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Old December 4th, 2008, 02:00 PM
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Originally Posted by Mathstud28 View Post
Ok then make it \sqrt{x^2+(-a)} and make the sub x=\sqrt{-a}\tan(\theta) or x=\sqrt{-a}\sinh(\theta)

but you took the square root of a neg number there. Is that allowed ?
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Old December 4th, 2008, 02:00 PM
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but you took the square root of a neg number there. Is that allowed ?
But you said what if a<0\implies{-a>0} so yes
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\int_0^z\ln\Gamma(x+1)dx=\frac{z}{2}\ln(2\pi)-\frac{z(z+1)}{2}+z\ln\Gamma(z+1)-\ln\left\{(2\pi)^{\frac{z}{2}}\exp\left(\frac{-z(z+1)}{2}-\frac{\gamma z^2}{2}\right)\prod_{k=1}^{\infty}\left\{\left(1+\frac{z}{k}\right)^k\exp\left(-z+\frac{z^2}{2k}\right)\right\}\right\}
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Old December 4th, 2008, 02:01 PM
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you're a very stupid man jhevon
And you're banned for 2 weeks to cool off. None of that.
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