Conference rusure::math

    what is   X^^4  means ? is it X to the power 4 to the power 4?
    /naser

        I think he meant "x to the fourth power" and mixed the
        FORTRAN notation x**4 with the notation often used here
        x^4 and came up with x^^4.
        
        One way to solve it would be to factor 1 + x^4 over the
        complex numbers and use partial fractions.
        
        Dan

    
    Hi All,
    
    Wouldn't it be better to use a substitution techneque.
    
    Let U = x(sqr) therefore dU = 2xdx --> dx = dU/2x
                   then replacing the 2x with 2U(power)0.5
    
    This would give us :      1             dU
                         ----------- X ------------
                          1 + U(sqr)   2U(power)0.5
    
    
    	This can then be integrated using partial Integration U.V,
    
    Would this method be feasable or is it a load of mathematical rubbish!

	it's a load of mathematical rubbish

    In closed form, this is familiar from electrical engineering, where
    one wants to find the noise bandwidth of an n'th order Butterworth
    lowpass filter.  The usual trick is via a contour integral...

    If you expanded the integrand in a series about 0 and integrated
    term by term you'd get

	F(x) = x - 1/5 x^5 + 1/9 x^9 - 1/13 x^13 + ...

    I don't see this as an obvious simple function; does anyone else?

    Looking at a "table of integrals" one finds

Vitus $ maple

    |\^/|
._|\|   |/|_. Licensed to Digital Equipment of Canada
 \  MAPLE  /  Version 4.3 --- Mar 1989
 <____ ____>  For on-line help, type  help();
      |

> int(1/(1+x^4),x);

             2      1/2
            x  + x 2    + 1                1/2                    1/2
         ln(---------------) + 2 arctan(x 2    + 1) + 2 arctan(x 2    - 1)
             2      1/2
            x  - x 2    + 1
     1/4 -----------------------------------------------------------------
                                        1/2
                                       2

> int(1/(1-x^4),x);
                         1/2 arctanh(x) + 1/2 arctan(x)

> quit;
bytes used=319132, alloc=147456, time=4.730

    The second form of integrand looks a lot simpler :-)

    Maple, when asked to simplify the integral merely reformatted it!

    - Jim [lazy button pusher today]

Could someone please show how to do this problem with contour integration ?

I learned it back in college, but in my 17 years at Digital I've never
needed it so I've forgotten it.

But I remember being quite impressed with how easy it was when I knew it.

/Eric

>> Could someone please show how to do this problem with contour integration ?
        
        The contour integral around a circle of the function z^n
        (assuming the origin is not on the circle when n is
        negative) is zero, except in the case where n = -1 and
        the origin is inside the circle, in which case the
        contour integral is `2 pi i'.
        
        Just generalize from there. :-)
        
        I don't think it will help in this problem, where an
        indefinite integral, and not a definite integral, is
        wanted.
        
        Dan

    To evaluate the definite integral of int(1/(1+z^n),z=0..infinity)
    it is easier to make a change of variables and use the reflection
    formula for the Gamma function.

    Let z = t^(1/n), then the integrand is

	1 t^(1/n-1)
	- -------- dt
	n  1 + t

    Which is integrates to

				      pi/n
	gamma(1/n) * gamma(1-1/n) = ---------
				    sin(pi/n)

    For the specific case of 1/(1+z^4), there are two poles above the
    real axis at .707 +/- .707 i.  The sum of their residues will equal
    1/(2 pi i)*integral, from which you can solve for the integral.

    The same trick applies with a different countour to establish the general
    reflection formula for the gamma function.  I'm a little rusty on the
    exact contour for that, but it's in any book on complex analysis,
    or a problem in the Schaums Outline...

    I don't know of an elegant algebraic trick for the indefinite integral
    though.

    For an k'th order butterworth filter, n = 2 k above.  The noise
    bandwidths of other filter shapes (Legendre, Chebyshev, or even
    1/3 octave bandpass ISO filters for audio noise measurement) are
    not so nice and it's easier to just numerically integrate.

    - Jim

	For the indefinite integral, you just split into partial
fractions surely, as Dan suggested in an early note.   The denominator
factors obviously into 4 linear factions, whose reciprocals can be integrated
as long as one is careful to consider what complex logarithms are
all about.   This is why the first of the Maple integrals that Jim performed
is more complicated than the second, where the coefficients of the integrands
are real or purely imaginary.

Regards,

	re .5,

>>             2      1/2
>>            x  + x 2    + 1                1/2                    1/2
>>         ln(---------------) + 2 arctan(x 2    + 1) + 2 arctan(x 2    - 1)
>>             2      1/2
>>            x  - x 2    + 1
>>     1/4 -----------------------------------------------------------------
>>                                        1/2
>>                                       2

	Actually that isn't too hard to derive. :-)   If you don't feel like
	dealing with the complex numbers that you get from writing 1 + x^4
	as a product of the four terms (x - (+/- 1 +/- i)/sqrt(2)), then
	make the following substitution:

		1 + x^4 = (1/4)(4 + 4x^4) = (1/4)(4 + u^4)

	In other words, 4x^4 = u^4 or u = x sqrt(2).  Why do this?  Because
	4 + u^4 factors as a difference of two squares, 4 + u^4 =
	(4 + 4u^2 + u^4) - 4u^2 = (2 + u^2)^2 - (2u)^2 =
	(2 + u^2 + 2u)(2 + u^2 - 2u) = (2 + 2u + u^2)(2 - 2u + u^2).

	So plug in this and dx sqrt(2) = du in order to get

	dx/(1 + x^4) = (du / sqrt(2)) / ((1/4)(4 + u^4))
		= (4/sqrt(2)) du / (4 + u^4)

	Now after factoring the denominator as above, we break it up using
	partial fractions into

		= (4/sqrt(2)) du (1/8) ((u+2)/(u^2+2u+2) - (u-2)/(u^2-2u+2))

		= (1/2sqrt(2)) du ( (u+1)/(u^2+2u+2)
				+       1/(u^2+2u+2)
				-   (u-1)/(u^2-2u+2)
				+       1/(u^2-2u+2) )

	Integrating these four is easy:

	indefinite integral of (u+1)/(u^2+2u+2) = (1/2) ln(u^2+2u+2) + c1
	indefinite integral of 1/(u^2+2u+2) = arctan(u+1) + c2
	indefinite integral of (u-1)/(u^2-2u+2) = (1/2) ln(u^2-2u+2) + c3
	indefinite integral of 1/(u^2-2u+2) = arctan(u-1) + c4

	Combining these gives the result

		       1    u^2+2u+2
	(1/2sqrt(2)) ( - ln -------- + arctan(u+1) + arctan(u-1) ) + c
		       2    u^2-2u+2

	which when you replace u = x sqrt(2) is what Maple gave.  You can
	also combine arctan(u+1) + arctan(u-1) by using the formula
	tan(A + B) = (tan A + tan B) / (1 - tan A tan B) with the result
	arctan(u+1) + arctan(u-1) = arctan( (u+1+u-1)/(1 - (u+1)(u-1)) )
	= arctan(2u/(2-u^2)).

	The other one is easier as 1 - x^4 factors into (1-x^2)(1+x^2)

	Dan

    
    Sorry for the original notation, I did mean "x to the fourth power".
    
    And here's a simple substitution scheme that helps reduce to
    a known standard form,
    
    
    
    The substitution that works algebraically is the
    following :
    
    	first, divide numerator and denominator by x**2, giving us
    
    	          (1/x**2)
    		---------------- ,
    		 (1/x**2) + x**2
    
    	rewrite this as,
    
    		   1   [(1/x**2) -1]         1   [(1/x**2) +1]  
    		  ---  ---------------  +   ---  ----------------
    		   2   [(x+1/x)**2 + 2]      2   [(x-1/x)**2 - 2]
    
    	substituing  x+(1/x) = t in the first integral, and
    	substituting x-(1/x) = u in the second integral,
    
    	we reduce the integral to known standard forms, I've
    	left out a few signs and the final few steps here, but 
    	the answer provided in one of the earlier replies
    	follows from these substitutions.            
                        	
    
    

    
    	whoops, the denominators in the 2 terms in the previous
    	reply should have been
    
    	[(1/x + x)**2 -2]     and [(x-1/x)**2 + 2] respectively,
    
    	sorry for the mistake ...

    Have I missed something or is not the obvious factoring of
    
         x^4+1=(x^2+2^(1/2)x+1)(x^2-2^(1/2)+1)
    
    then proceeding with partial fractions and simple completion of
    of the squares not the way to go?  The four terms in the partial
    fraction expansion divide up into two arctangents in the result
    and two logarithms in the result.  This seems mechanical but
    rational functions are about the most mechanical integral there
    is.  Of slight interest is the famous theorem proved very
    mechanically that rational functions can always be integrated
    as elementary functions (sometimes refered to as closed form -
    I don't understand why) giving nothing nastier than logs and
    arctangents.
    
    Chuck

        re .13,
        
        That's essentially what was in .10, with a substitution
        to keep the sqrt(2) out of the way for a while.
        
        Dan