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From: slawek on 18 Jul 2010 04:43

U�ytkownik "karl" <oudeis(a)nononet.com> napisa� w wiadomo�ci grup
dyskusyjnych:4c4292de$0$7654$9b4e6d93(a)newsspool1.arcor-online.net...
>>> But in the functions you have to use the substition: t=cx everywhere.
>>> Try it with a simple example
>> O.K., we take f(x)= x.

A good idea anyway.

I = Integrate[Exp[-2 x] f(x) , {x, xa, xb} ]
where
f(x) = x
xa = 0
xb = Infinity

gives I = 1/4

http://www.wolframalpha.com/input/?i=Integrate[+x+Exp[-2+x]+,+{x,+0,+Infinity}+]+



I = g * Integrate[Exp[-t] F[t], {t, ta, tb}]

t = c x = 2 x, i.e. c = 2

ta = c xa = 2 xa = 2 * 0 = 0
tb = c xb = 2 xb = 2 * Infinity = 0

g = 1/c = 1/2

F[t] = f[x] and f[x] = x thus F[t] = t/2 because x = t/2

Therefore

I = 1/2 Integrate[t/2 Exp[-t], {t, 0, Infinity}]
gives I = 1/4

http://www.wolframalpha.com/input/?i=1/2*Integrate[t/2+Exp[-t],+{t,+0,+Infinity}]


I use calculus for 30 years. I am pretty sure that I am not an infallible
god. But I think I'm right this time and these calculations are correct.



From: karl on 18 Jul 2010 08:59
Am 18.07.2010 07:34, schrieb karl:
> Am 18.07.2010 07:32, schrieb karl:
>> Am 17.07.2010 20:50, schrieb karl:
>>> Am 17.07.2010 15:52, schrieb slawek:
>>>>
>>>> U�ytkownik "karl" <oudeis(a)nononet.com> napisa� w wiadomo�ci grup
>>>> dyskusyjnych:4c41a8ff$0$6975$9b4e6d93(a)newsspool4.arcor-online.net...
>>>>> Hi, in the step from x as variable to t as variable I do not see how
>>>>> you get F[t].
>>>>
>>>> It doesn't matter on this level of the abstraction. You are right, it
>>>> would be possible to derieve formula for F from the (known) f function.
>>>>
>>>>> How can you justify that you don`t replace x by t and not by t/c?
>>>>
>>>> Jacobian is a constant and can be pulled out as g.
>>>>
>>>> It may be written d(t/c) in the integral, but usually it is written as
>>>> (1/c) dt.
>>>> .
>>>
>>> But in the functions you have to use the substition: t=cx everywhere.
>>> Try it with a simple example
>>>
>>
>>
>> O.K., we take f(x)= x.
>>
>> I = Integrate[Exp[-c x] x ]
>>
>>
>> and as limits 0 and + oo, then we have for the integral:
>>
>> I = 1/c^2
>>
>> (see Abramowitz/Stegun, p.71, eq. 4.2.55)
>>
>> Substitution: t=cx. You say this gives
>>
>> I =(1/c) Integrate[Exp[-t] t ]= 2/c.
>
> There is a typo, should be:
>
> I =(1/c) Integrate[Exp[-t] t ]= 1/c.
>
Also something is unclear with the limits. Should this mean

I = Integrate[Exp[-c x] f[x], {x, xa, xb}]
I = g * Integrate[Exp[-t] F[t], {t, ta, tb}]

that the lower limit is xa and the upper xb and in the second case ta and tb? But x and t are the integration variables
in both cases!







From: slawek on 18 Jul 2010 10:01

U�ytkownik "karl" <oudeis(a)nononet.com> napisa� w wiadomo�ci grup
dyskusyjnych:4c42fb02$0$6761$9b4e6d93(a)newsspool3.arcor-online.net...
> that the lower limit is xa and the upper xb and in the second case ta and
> tb? But x and t are the integration variables
> in both cases!

xa means x-subscript-a, in LaTex it would be written as x_a, in Mathematica
x a != xa, i.e. "xa" is not a product of "a" and "x".

ta means t-subscript-b, x_b



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