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Integration of $\displaystyle \int\frac{x^2-1}{\sqrt{x^4+1}} \,dx$

$\bf{My\; Try}$:: Let $x^2=\tan \theta$ and $\displaystyle 2xdx = \sec^2 \theta \, d\theta\Rightarrow dx = \frac{\sec^2 \theta}{2\sqrt{\tan \theta}} \, d\theta$

$$ \begin{align} & = \int\frac{\tan \theta - 1}{\sec \theta}\cdot \frac{\sec^2 \theta}{2\sqrt{\tan \theta}} \, d\theta = \frac{1}{2}\int \frac{\left(\tan \theta - 1\right)\cdot \sec \theta}{\sqrt{\tan \theta}} \, d\theta \\ & = \frac{1}{2}\int \left(\sqrt{\tan \theta}-\sqrt{\cot \theta}\right)\cdot \sec \theta \, d\theta \end{align} $$

Now i did not understand how can i solve it

Help me

Thanks

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  • $\begingroup$ Are you sure that this is doable? $\endgroup$ – imranfat Nov 7 '13 at 16:49
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    $\begingroup$ See wolfram. It's not pretty. $\endgroup$ – Namaste Nov 7 '13 at 16:53
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For any real number of $x$ ,

When $|x|\leq1$ ,

$\int\dfrac{x^2-1}{\sqrt{x^4+1}}dx$

$=\int(x^2-1)\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{4n}}{4^n(n!)^2}dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{4n+2}}{4^n(n!)^2}dx-\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{4n}}{4^n(n!)^2}dx$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{4n+3}}{4^n(n!)^2(4n+3)}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{4n+1}}{4^n(n!)^2(4n+1)}+C$

When $|x|\geq1$ ,

$\int\dfrac{x^2-1}{\sqrt{x^4+1}}dx$

$=\int\dfrac{x^2-1}{x^2\sqrt{1+\dfrac{1}{x^4}}}dx$

$=\int\left(1-\dfrac{1}{x^2}\right)\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{-4n}}{4^n(n!)^2}dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{-4n}}{4^n(n!)^2}dx-\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{-4n-2}}{4^n(n!)^2}dx$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{1-4n}}{4^n(n!)^2(1-4n)}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{-4n-1}}{4^n(n!)^2(-4n-1)}+C$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(4n+1)x^{4n+1}}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(4n-1)x^{4n-1}}+C$

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As per the solution I derived here:

\begin{align} I(x,a,k,n,m) &=\int_0^x \frac{t^k}{\left(t^n + a\right)^m}\:dt\\ & = \frac{1}{n}a^{\frac{k + 1}{n} - m} \left[B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right) - B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}, \frac{1}{1 + ax^n} \right)\right] \end{align}

We position your integral as:

\begin{align} I &= \int \frac{x^2 - 1}{\sqrt{x^4 + 1}}\:dx = \int_{0}^{x} \frac{t^2}{\sqrt{t^4 + 1}}\:dt = \int_{0}^{x} \frac{t^2 - 1}{\sqrt{t^4 + 1}}\:dt - \int_{0}^{x} \frac{1}{\sqrt{t^4 + 1}}\:dt \\ &= I\left(x,1,2,4, \frac{1}{2}\right) - I\left(x,1,0,4, \frac{1}{2}\right) \\ &= \frac{1}{4}1^{\frac{2 + 1}{4} - \frac{1}{2}} \left[B\left(\frac{1}{2} - \frac{2 + 1}{4}, \frac{2 + 1}{4}\right) - B\left(\frac{1}{2} - \frac{2 + 1}{4}, \frac{2 + 1}{4}, \frac{1}{1 + 1\cdot x^4} \right)\right] - \frac{1}{4}1^{\frac{0 + 1}{4} - \frac{1}{2}} \left[B\left(\frac{1}{2} - \frac{0 + 1}{4}, \frac{0 + 1}{4}\right) - B\left(\frac{1}{2} - \frac{0 + 1}{4}, \frac{0 + 1}{4}, \frac{1}{1 + 1\cdot x^4} \right)\right] \\ &= \frac{1}{4}\left[B\left(-\frac{1}{4}, \frac{3}{4}\right) - B\left(-\frac{1}{4}, \frac{3}{4}, \frac{1}{1 + x^4}\right) \right] - \frac{1}{4}\left[B\left(\frac{1}{4}, \frac{1}{4}\right) - B\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{1 + x^4}\right) \right] \end{align}

Using the relationship between the Beta and Gamma Functions, we have:

\begin{align} B\left(-\frac{1}{4}, \frac{3}{4}\right) &=\frac{\Gamma\left(-\frac{1}{4}\right)\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(-\frac{1}{4} + \frac{3}{4}\right)} & B\left(\frac{1}{4}, \frac{1}{4}\right)&=\frac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{1}{4} + \frac{1}{4}\right)} \\ &= \frac{\Gamma\left(-\frac{1}{4}\right)\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{2}\right)} & &= \frac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{2} \right)} \\ &= \frac{\Gamma\left(-\frac{1}{4}\right)\Gamma\left(\frac{3}{4}\right)}{\sqrt{\pi}} & &= \frac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{3}{4}\right)}{\sqrt{\pi}} \end{align}

Thus:

\begin{align} I &= \frac{1}{4}\left[B\left(-\frac{1}{4}, \frac{3}{4}\right) - B\left(-\frac{1}{4}, \frac{3}{4}, \frac{1}{1 + x^4}\right) \right] - \frac{1}{4}\left[B\left(\frac{1}{4}, \frac{1}{4}\right) - B\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{1 + x^4}\right) \right] \\ &= \frac{1}{4}\left[B\left(-\frac{1}{4}, \frac{3}{4}\right) - B\left(\frac{1}{4}, \frac{1}{4}\right) \right] + \frac{1}{4}\left[B\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{1 + x^4}\right) - B\left(-\frac{1}{4}, \frac{3}{4}, \frac{1}{1 + x^4}\right) \right] \\ &= \frac{\Gamma\left(\frac{3}{4}\right)}{4\sqrt{\pi}}\left[ \Gamma\left(-\frac{1}{4}\right) - \Gamma\left(\frac{1}{4}\right)\right]+ \frac{1}{4}\left[B\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{1 + x^4}\right) - B\left(-\frac{1}{4}, \frac{3}{4}, \frac{1}{1 + x^4}\right) \right] \end{align}

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    $\begingroup$ Look at you, using your own special function! Nice work as usual David $\endgroup$ – clathratus Jan 2 at 2:39
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    $\begingroup$ Haha, going by the suggested question list I could spend the next 6 months with this alone! $\endgroup$ – user150203 Jan 2 at 2:40
  • $\begingroup$ There is a question that comes out of this. Is there anything that can be used to simplify $\Gamma(−z)−\Gamma(z)$? $\endgroup$ – user150203 Jan 2 at 2:53
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    $\begingroup$ I do not know if this "simplifies" things, but it is an alternate form. Recall $$\Gamma(s)=\frac{\Gamma(s+1)}{s}$$ Replace $s$ with $-s$ to get $$\Gamma(-s)=-\frac1s\Gamma(1-s)$$ Then use the $\Gamma$ reflection formula to see that $$\Gamma(-s)=\frac{-\pi}{s\Gamma(s)\sin\pi s}=\frac{-\pi}{\Gamma(s+1)\sin\pi s}$$ $\endgroup$ – clathratus Jan 2 at 4:52
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Binomial expansion of the expression to used in the respective integrals

Use binomial expansions of the expressions in respective integrals as given in the previous solution of the problem.

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  • $\begingroup$ Can you make this more readable and easier on the eyes? Perhaps typing out your answer may help. $\endgroup$ – JavaMan Jan 2 at 3:42

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