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- Jan 26, 2012

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This does not have an integral in terms of elementary functions by the theorem of Chebyshev that you have had pointed at in an earlier post.$\displaystyle \int\frac{1}{\sqrt[4]{1-x^4}}dx$

It can also be found >>here<< near expression (20)

CB

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- Jan 26, 2012

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A quicker way to Wolfram straight from MHB is using one of our MHB Widgets.This does not have an integral in terms of elementary functions by the theorem of Chebyshev that you have had pointed at in an earlier post.

It can also be found >>here<< near expression (20)

CB

For your question this will use the same Wolfram indefinite integral calculator. Here's the output of your question.

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- Feb 7, 2012

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- Feb 13, 2012

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According to...Notices of the American Math. Soc.

Gauss's Constant -- from Wolfram MathWorld

... the so called 'Gauss's constant' is defined as...

$\displaystyle G= \frac{1}{\text{M}\ (1, \sqrt{2})} = \frac{2}{\pi} \int_{0}^{1} \frac{dx}{\sqrt{1-x^{4}}}$

Kind regards

$\chi$ $\sigma$

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(Dont no whether it is Right or not)

$\displaystyle \int\frac{1}{\sqrt[4]{1-x^4}}dx$

Put $\displaystyle 1-x^4=t^4x^4\Leftrightarrow x^4 = \frac{1}{1+t^4}$

$\displaystyle4x^3 dx = -\frac{4t^3}{(1+t^4)^2}dt$

$\displaystyle dx = -\frac{t^3}{(1+t^4)^2.x^3}dt$

So Integral Convert into $\displaystyle -\int\frac{t^3}{(1+t^4)^2.x^3.tx}$

$\displaystyle - \int\frac{t^2}{(1+t^4)^2}.\frac{(1+t^4)}{1}dt$

$\displaystyle -\int\frac{t^2}{1+t^4}dt = -\frac{1}{2}\int\frac{(t^2+1)+(t^2-1)}{1+t^4}dt$

Now Let $\displaystyle \mathbb{I = \int \frac{t^2+1}{t^4+1}}$ and $\displaystyle\mathbb{J=\int\frac{t^2-1}{t^4+1}}dt$

after that we can solve for $\mathbb{I}$ and $\mathbb{J}$

Thanks

- Feb 13, 2012

- 1,704

Very good!... pratically speacking You can directly solve the integral...

(Dont no whether it is Right or not)

$\displaystyle \int\frac{1}{\sqrt[4]{1-x^4}}dx$

Put $\displaystyle 1-x^4=t^4x^4\Leftrightarrow x^4 = \frac{1}{1+t^4}$

$\displaystyle4x^3 dx = -\frac{4t^3}{(1+t^4)^2}dt$

$\displaystyle dx = -\frac{t^3}{(1+t^4)^2.x^3}dt$

So Integral Convert into $\displaystyle -\int\frac{t^3}{(1+t^4)^2.x^3.tx}$

$\displaystyle - \int\frac{t^2}{(1+t^4)^2}.\frac{(1+t^4)}{1}dt$

$\displaystyle -\int\frac{t^2}{1+t^4}dt = -\frac{1}{2}\int\frac{(t^2+1)+(t^2-1)}{1+t^4}dt$

Now Let $\displaystyle \mathbb{I = \int \frac{t^2+1}{t^4+1}}$ and $\displaystyle\mathbb{J=\int\frac{t^2-1}{t^4+1}}dt$

after that we can solve for $\mathbb{I}$ and $\mathbb{J}$

Thanks

$\displaystyle - \int\frac{t^2}{1+t^4}dt = \frac{1}{4\ \sqrt{2}}\ [\ln (t^{2}+ \sqrt{2}\ t +1) - \ln (t^{2} - \sqrt{2}\ t +1) + 2\ \tan^{-1} (1+\sqrt{2}\ t) - 2\ \tan^{-1} (1- \sqrt{2}\ t)] + c$

Kind regards

$\chi$ $\sigma$

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- Feb 13, 2012

- 1,704

From...This does not have an integral in terms of elementary functions by the theorem of Chebyshev that you have had pointed at in an earlier post.

It can also be found >>here<< near expression (20)

CB

Indefinite Integral -- from Wolfram MathWorld

$\displaystyle \int x^{u}\ (a + b\ x^{v})^{w}$ (1)

...

In this case is $w=-\frac{1}{4}$, $u=0$ and $v=4$ so that...

$\displaystyle w+\frac{u+1}{v}=0$ (2)

... and, if I remember well, $0 \in \mathbb{Z}$...

Kind regards

$\chi$ $\sigma$

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- #9

- Feb 13, 2012

- 1,704

You can expand the function under integral sign in partial fractions...To Chisigma

How can I calculate it Directily

$\displaystyle - \int\frac{t^2}{1+t^4}dt = \frac{1}{4\ \sqrt{2}}\ [\ln (t^{2}+ \sqrt{2}\ t +1) - \ln (t^{2} - \sqrt{2}\ t +1) + 2\ \tan^{-1} (1+\sqrt{2}\ t) - 2\ \tan^{-1} (1- \sqrt{2}\ t)] + c$

Thanks

$\displaystyle f(t)=\frac{t^{2}}{1+t^{4}}= \frac{r_{1}}{t-\sqrt{2}\ (1+i)} + \frac{r_{2}}{t-\sqrt{2}\ (1-i)} + \frac{r_{3}}{t+\sqrt{2}\ (1+i)} + \frac{r_{4}}{t+\sqrt{2}\ (1-i)}$ (1)

... where...

$\displaystyle r_{1}= \lim_{t \rightarrow \sqrt{2}\ (1+i)} (t-\sqrt{2}\ (1+i))\ f(t)$

$\displaystyle r_{2}= \lim_{t \rightarrow \sqrt{2}\ (1-i)} (t-\sqrt{2}\ (1-i))\ f(t)$

$\displaystyle r_{3}= \lim_{t \rightarrow - \sqrt{2}\ (1+i)} (t+\sqrt{2}\ (1+i))\ f(t)$

$\displaystyle r_{4}= \lim_{t \rightarrow -\sqrt{2}\ (1-i)} (t+\sqrt{2}\ (1-i))\ f(t)$ (4)

... and then integrate 'term by term'...

Kind regards

$\chi$ $\sigma$

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