1. We see at right that the sum and the integral differ by the triangular shaped notches near the curve. Let us estimate the area of the difference of the jth column, D, as
   
   D_j = ^j_j-1 exp(-²x²) dx - exp(-²x²)
   
   That integral is the error function which has no closed form solution except for which, in fact, we're to evaluate in the 2nd part of this problem!
   
   So we are going to take literally the suggestion that the D_j are triangular and use a triangular area to estimate the difference. Since the base of each column (and triangle) is just one, the triangular area is equal to ½ the height of the triangle, or
   
   D_j = ½ [ exp(-²(j-1)²) - exp(-²j²) ]
   
   Expanding (j-1)² gives (j²-2j+1) as additive terms in an the exponential, but we know that e^x±y=e^xe^±y, we can factor out exp(-²j²) leaving
   
   D_j = ½ exp(-²j²) * [ exp(+2²j-²) - 1 ]
   
   Noting that for vanishingly small argument e1, we can make the term in [...] vanish in the limit that ²0.
   
   So it's near-zero values of that do the trick!
   
   
2. That dxdy elemental area in polar coordinates becomes dr, the small extension of a radial arm, times rd, the small arc length through which d sweeps the arm. And since x and y are always in the first quadrant (it's after all), that means that ranges from 0 to /2 only. The radius takes on all positive values; negative radii are meaningless.
   
   Since x and y are independent (as are r and ), we can take them in and out of one another's integrals at will. So
   
   I = exp(-²x²)dx = exp(-²y²)dy
   
   then
   
   I ² = exp[-²(x²+y²)] dx dy
   
   I ² = ₀^/2 exp[-²r²] rdr d
   
   And if we notice that the derivative of that exponential with respect to r is just
   
   -2r² exp[-²r²],
   
   then the integrand becomes nothing more than
   
   r exp(-²r²) = -(1/2²) d [exp(-²r²)] / dr
   
   And the advantage for us is that the integral of a derivative is the derivative's function evaluated at the endpoints!
   
   r exp(-²r²) dr = - (1/2²) [exp(-²r²) | ₀ = - (1/2²) (0-1) = + (1/2²)
   
   The other integral is trivial:
   
   
   it is ₀^/2 d = /2
   
   making the double integral
   
   I ² = (1/2²)(/2) = /4²
   
   or
   
   I = ()^½/2
   
   as advertized.
   

Questions?

Chris Parr
University of Texas at Dallas
Programs in Chemistry, Room Berkner 3.506
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Richardson, TX  75083-0688

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