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Old 12-02-2008, 08:24 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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Old 12-02-2008, 09:05 AM
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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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Old 12-02-2008, 09:39 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 12-04-2008, 02:53 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; 12-04-2008 at 09:21 PM.
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Old 12-04-2008, 02:56 PM
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mathstud what if a is negative - then we can't consider sqrt(a) ?
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Old 12-04-2008, 02:57 PM
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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 12-04-2008, 02:59 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 12-04-2008, 03:00 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 12-04-2008, 03:00 PM
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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 12-04-2008, 03:00 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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