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Indefinite Integral

jacks

Well-known member
Apr 5, 2012
226
$\displaystyle \int\frac{1}{\sqrt[4]{1-x^4}}dx$
 

CaptainBlack

Well-known member
Jan 26, 2012
890
$\displaystyle \int\frac{1}{\sqrt[4]{1-x^4}}dx$
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
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
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
A quicker way to Wolfram straight from MHB is using one of our MHB Widgets.

Screen Shot 2012-10-15 at 2.55.57 AM.png

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

Screen Shot 2012-10-15 at 2.55.19 AM.png
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
There is a neat formula (discovered by Gauss) for the definite integral of $1/ \sqrt[4]{1-x^4}$, namely $$\int_0^1 \frac{1}{\sqrt[4]{1-x^4}}dx = \frac{\pi}{2M(\sqrt2)},$$ where $M(x)$ denotes the arithmetic-geometric mean of 1 and $x$. See the very interesting article An Eloquent Formula for the Perimeter of an Ellipse in last month's issue of the Notices of the American Math. Soc.
 

chisigma

Well-known member
Feb 13, 2012
1,704
There is a neat formula (discovered by Gauss) for the definite integral of $1/ \sqrt[4]{1-x^4}$, namely $$\int_0^1 \frac{1}{\sqrt[4]{1-x^4}}dx = \frac{\pi}{2M(\sqrt2)},$$ where $M(x)$ denotes the arithmetic-geometric mean of 1 and $x$. See the very interesting article An Eloquent Formula for the Perimeter of an Ellipse in last month's issue of the Notices of the American Math. Soc.
According to...

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

$\chi$ $\sigma$
 

jacks

Well-known member
Apr 5, 2012
226
But I have solved like that way.

(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
 

chisigma

Well-known member
Feb 13, 2012
1,704
But I have solved like that way.

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

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

Kind regards

$\chi$ $\sigma$
 
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chisigma

Well-known member
Feb 13, 2012
1,704
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
From...

Indefinite Integral -- from Wolfram MathWorld

Chebyshev proved that if $u$,$v$ and $w$ are rational numbers, then...


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


...is integrable in terms of elementary functions iff $\frac{u+1}{v}$, $w$ or $w+\frac{u+1}{v}$ is an integer...


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

$\chi$ $\sigma$
 

jacks

Well-known member
Apr 5, 2012
226
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
 

chisigma

Well-known member
Feb 13, 2012
1,704
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
You can expand the function under integral sign in partial fractions...

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