Consider

f  z    exp  2tz  t 2  dt
0

f   z    exp  2tz  t 2  dt
0
Let t = -t
f z  
0
 exp  2tz  t  dt
2

So that f  z   f   z  

 exp  2tz  t  dt
2

t 2  2tz  16 * 3  0
b
1
t

b 2  4ac
2a 2a
2z 1
t

4 z 2  64 * 3
2 2
 16 * 3  z 0 16 * 3
tmax  z  z 2  16 * 3  z  z 1 
 
2z
2z2 

z
-1
-4
-8
-64
1
2
4
tmax
3.12
1.66
0.94
0.12
5.12
6.47
9.65
Plotswere made using tbliFbrack.zip
g  t   exp  2tz  t 2 
g
 2  z  t  exp  2tz  t 2 
t
2 g
2
 2t  4  z  t  exp  2tz  t 2 
2
t
  4 z 2  10 zt  4t 2  exp  2tz  t 2 

3 g

t 3

 10 z  8t   2  z  t   4 z
2

 10 zt  4t 2  exp  2tz  t 2 
 10 z  8t  2 z 3  20 z 2t  8 zt 2 
2
 
 exp  2tz  t 
2
2
3
 8tz  20 zt  8t



 10 z  2 z 3   8t  28 z 2t  28 zt 2  8t 3  exp  2tz  t 2 
At t=0
g 
 2z
t  t 0
3 g 
  10 z  2 z 3 
3 
t  t 0
The integrations will be made using End point trap rule.docx.
These integrations include
h4
h4
f
'''(
end
)

f
'''(
beg
)



f '''(beg )


720
720
So the first error term is approximately
h6 4  5
h6 3
f '''(beg ) 
z
720
18
Richardson’s extrapolation will be made by comparing the result with 2hpoints to that with h points,
making the final error err  7h8 z 3


z  4



While the integral is approximately 0.5  tmax  0.5  z  z 1  16 / z 2 
 0.5   z  z 
The h required for machine accuracy is
For z << -4, this becomes h8 
h  102 /
err
 1016 
integral
8
z



7 h8 z 3

8
0.5   z  z  
z

4
1016
4
7z
z
z 4

In this same large negative z limit tmax 
So that the number of intervals becomes N p 
8
z
8 z
z 10
2

800
z
This is on the order of 60 for z = - 100 which is the number used in theplots above.
In general Np will be a factor of 6 so that the integral can be evaluated with all point, ½ points and then
1/3 points
These points will be fitted to a  bh 6 . The final error in the integral is the standard deviation in a.