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From: karl on 17 Jul 2010 08:56
Am 16.07.2010 19:38, schrieb slawek:
> This question is really stupid.
>
> We have got an integral (where xa,xb,c etc. are constants; I use a
> pseudo-Mathematica notation for the clarity)
>
> I = Integrate[Exp[-c x] f[x], {x, xa, xb}]
>
> we want to obtain the form
>
> I = g * Integrate[Exp[-t] F[t], {t, ta, tb}]
>
> Obviously,
>
> t == c x
> dt == c dx
>
> ta == c xa
> tb == c xb
>
> g == 1/c
> F[t] == f[x]
>
>
> Maybe I miss something? Well, all variables are real.
>
>
>
Hi, in the step from x as variable to t as variable I do not see how you get F[t]. If
we have t=cx and x=t/c, I would assume that the second form is:

I = g * Integrate[Exp[-t] f[t/c], {t/c, ta/c, tb/c}]

How can you justify that you don`t replace x by t and not by t/c?

Ciao

Karl
From: slawek on 17 Jul 2010 09:52

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.





From: karl on 17 Jul 2010 14:50
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

From: karl on 18 Jul 2010 01:32
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.

So your result is wrong.

Ciao

Karl
From: karl on 18 Jul 2010 01:34
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.

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