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Old 11-19-2008, 08:50 AM
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Default Changing the order of integration
Hey I'm having some trouble with multiple integrals if anyone would care to explain (I think I'm just not using the correct region!)

Quote:
Evaluate
\int_{x=0}^2\int_{y=\frac{x}{2}}^1 2xy^2 dy dx

first directly then changing the order of integration
I did the first part and got 0.8 as the answer

just for the second I'm having trouble

I think the surface being integrated over the the triangle bounded by the lines y=x/2 x=0 and y=1 but I may be wrong which gives

\int_{y=0}^1\int_{x=2y}^2 2xy^2 dx dy

but the value of that doesnt agree, can anyone explain how to do the limits correctly?

cheers

Simon
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Old 11-19-2008, 10:19 AM
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\int\limits_0^1 {\int\limits_0^{x = 2y} {2xy^2 dxdy} }
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Old 11-19-2008, 02:53 PM
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Quote:
Originally Posted by thelostchild View Post
Hey I'm having some trouble with multiple integrals if anyone would care to explain (I think I'm just not using the correct region!)



I did the first part and got 0.8 as the answer

just for the second I'm having trouble

I think the surface being integrated over the the triangle bounded by the lines y=x/2 x=0 and y=1 but I may be wrong which gives

\int_{y=0}^1\int_{x=2y}^2 2xy^2 dx dy

but the value of that doesnt agree, can anyone explain how to do the limits correctly?

cheers

Simon
0\leqslant{x}\leqslant{2} and \frac{x}{2}\leqslant{y}\leqslant{1}\implies{x\leqslant{2y}\leqslant{2}}. Stringing these two inequalites together gives

0\leqslant{x}\leqslant{2y}\leqslant{2}

From there Peritus's answer should be more apparent.
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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 11-19-2008, 03:27 PM
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Quote:
Originally Posted by Mathstud28 View Post

0\leqslant{x}\leqslant{2} and \frac{x}{2}\leqslant{y}\leqslant{1}\implies{x\leqslant{2y}\leqslant{2}}. Stringing these two inequalites together gives

0\leqslant{x}\leqslant{2y}\leqslant{2}

From there Peritus's answer should be more apparent.
But be careful 'cause, this method not always works.
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Old 11-19-2008, 05:25 PM
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Originally Posted by Krizalid View Post
But be careful 'cause, this method not always works.
Really? I have never encountered a case where it hasn't. Would you mind giving me an example of one please?
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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 11-19-2008, 05:28 PM
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Try it with \int_0^1\int_{x^3}^{\sqrt[3]x}dy\,dx.
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Old 11-19-2008, 05:31 PM
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Quote:
Originally Posted by Krizalid View Post
Try it with \int_0^1\int_{x^3}^{\sqrt[3]x}dy\,dx.
Thank you, I will report back later when I have had time to look at it.
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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 11-20-2008, 11:30 AM
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Also be careful that reversing integration order is submitted to the condition that \int \int \left|f(x,y)\right| ~ dx ~ dy is a finite value.
See Fubini's theorem - Wikipedia, the free encyclopedia for further information (advanced calculus imo)
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