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Let $I_{n}(x)=\int (x^{2}+1)^{(n-0.5)} dx$. I want to express $I_{n}(x)$ with earlier terms $\sum_{k\leq n} h_{k} I_{k}(x)$ where $h_{k}$ is a constant. I have no idea by which terms.

If I am going to do this manually, I am suspecting that I need to do here some trigonometric tricks to get this formula into some nice and easy form and then just simple integration by parts (here).

Trial 1. Dead end. Perhaps, useful $$\int \frac{1}{1+x^{2}} dx = Artcan(x) +C$$.

Trial 2. Dead end.

Let $II_{n}(x)=\int cos^{n}(x) dx$ so by a few times integration by parts:

$$n II_{n}(x)-(n-1)II_{n-2}(x)=sin(x)cos^{n-1}(x)$$,

this may become useful -- no idea, look I find dead end here.

Trial 3 and question. Some computational method?

I would like to know whether some program can help with this kind of problems. I find it very erroneous to do many integration-by-parts again-and-again, some computer software to check calculations or help with this kind of repetitive problems that can be very time-consuming to realize?

hhh
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2 Answers2

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Substituting $$x:=\sinh y\ ,\quad dx=\cosh y\ dy$$ you get $$I_n(x)=\int\cosh^{2n}y\ dy\Bigr|_{y:={\rm arsinh} x}\ .$$ Now you have to set up a recursion for the integrals $$J_n(y):=\int\cosh^{2n}y\ dy\ .$$ This can be done using partial integration: $$\eqalign{J_n(y)&=\int\cosh^{2n-1}y\ \cosh y\ dy\cr &= \cosh^{2n-1}y\ \sinh y - (2n-1)\int \cosh^{2n-2}y\ \sinh^2 y\ dy \cr &= \cosh^{2n-1}y\ \sinh y - (2n-1)\int \cosh^{2n-2}y\ (\cosh^2 y-1)\ dy\cr &=\cosh^{2n-1}y\ \sinh y -(2n-1)\bigl(J_n(y)-J_{n-1}(y)\bigr)\ .\cr }$$ This gives $$J_n(y)={1\over 2n}\bigl(\cosh^{2n-1}y\ \sinh y +(2n-1) J_{n-1}(y)\bigr)\ .$$ Finally the back-substitution: $\sinh y:=x$, $\ \cosh y:=\sqrt{1+x^2}$.

Christian Blatter
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  • Is it possible to try difference choices for integration by parts with some software? Without having an idea for the choice of the terms, I find it very slow to try different combinations. How can I know that this equation for example has only one possible recursive form? – hhh Jan 23 '12 at 16:58
  • @hhh: See my edit. – Christian Blatter Jan 23 '12 at 21:20
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Plugging the integral into Mathematica gives

$$I_n(x) = x\ {}_2F_1\bigl(0.5, 0.5 - n; 1.5; -x^2\bigr)$$

The definition of the hypergeometric function ${}_2 F_1$ is well explained on Wikipedia, including a set of recurrence relations. The one of interest to you, in Wikipedia's notation, is

$$F(b-) = \frac{2b - c + (a-b)z}{b - c} F + \frac{1 - z}{c - b}F(b+)$$

You get this by equating the second and fifth lines of Gauss's contiguous relations in the Wikipedia article. Plugging in your values, if I've done the algebra right, you get

$$I(n) = \frac{0.5 + 2n + nx^2}{1+n}I(n-1) + \frac{1 + x^2}{1 + n}I(n-2)$$

Hypergeometric functions are well studied and many numeric libraries include methods to calculate them reasonably efficiently.

David Z
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  • Is there some easy way to replicate this process by hand or symbolically every step? I cannot see how the function can be expressed by such simple form. Now there is something missing, investigating. – hhh Jan 23 '12 at 17:08
  • I would try either converting your integral into one of the integral definitions of the hypergeometric function, or doing a series expansion within the integral and integrating it term by term. – David Z Jan 23 '12 at 19:00