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How to get the decision of integral in a symbolic form?

jjjjvb
1-Newbie

How to get the decision of integral in a symbolic form?

Mathematica gives an answer for indefinite integral. I got expression in the second line after the expansion of the limits. But I can`t plot with this expression. The plot appears for expression in the third line. Is it possible to get the decision of this integral any other way? Thank you.

13 REPLIES 13

According to Mathcad's numerical calculations the three expressions you have listed are all different! See attached.

You have three unspecified variables. Which do you want to plot against?

Why do you need a symbolic result if you just want plots - you could plot using numerics.

Alan

Thank for your answer. I need decision of integral in first line. A,b,c are constant and independent from variable x. I need plot result of this integral. So I can get result in simbolic form.

The integral in your file goes from 0 to infinity, so it has a single value. Do you mean the upper limit to be some finite value, x, and you want to plot the value of the integral as a function of this upper limit?

Alan

The primitive which iwill be got as a result of integration will be plotted from variable t which is included in c.

Then make c a function of t and simply plot the numerical result of the integral against t. If you post the values of a and b and the functional relationship between c and t, with the range of t-values of interest, I'll see if I can plot it.

Here is an example using arbitrary values

integ.PNG

Alan

a=8.5*10^11, b=8*10^8, c=0.4-8*10^5*t. Range t is like this: -5*10^-9...+10^-6.

How does Mathematica give the primitive of this integral in symbolic form?

jvm jjjvb wrote:

How does Mathematica give the primitive of this integral in symbolic form?

You mean Mathcad, and it doesn't! Numerical results below:

integ2.PNG

Alan

I`d like c(t) to have discrete time as it is shown in the attached file. Which changes is it necessary to make?

There aren't any discrete times listed in this file! What are the times of interest?

Alan

I mean value k+Tp where k=-255...+255. I wrote it in the file integ1c.

jvm jjjvb wrote:

I mean value k+Tp where k=-255...+255. I wrote it in the file integ1c.

Sorry, I looked at the wrong file!

However (with the right file) your function u1(a1,t) gives the same number for all your values of t to the accuracy of the numerics in Mathcad. With your original, unscaled function the value is always zero. With the function scaled so that x is replaced by y/a1, say, the value is always about 6.8*10^(-13) (identical for all values of t to 17 significant figures).

You need either to find a program that will provide much greater precision.(in a reasonable time - Mathcad's symbolic evaluation of each t takes a lifetime, even at its default 20 significant figures) or look for a fundamental reformulation of your problem that doesn't require such high precision.

Alan

Can I change the accuracy of calculation in setting, by analogy with zero threshold ?

jvm jjjvb wrote:

Can I change the accuracy of calculation in setting, by analogy with zero threshold ?

I don't think so.

If you look at the individual values of the summation over k within u1, you find that they are all effectively zero, except for the one term when t = k*Tp. This is true for all values of t and k. Thus, because, in this case, u is independent of t and k, you get the same result for all t.

Alan

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