Definite integral with complex number

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Pilar Jiménez 2017-1-17

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评论： Star Strider 2017-1-18

I need to solve an integral of the function f=(wn.*t).*exp(1i.*2.*pi.*t) where wn=1 and have complex numbers, it is a definite integral of 0 to 0.3 and I have to obtain a numeric answer. I tried to do with the comands int and integral but doesn`t help me because this comands uses symbolics variables and I need numeric answers, and I tried to obtain the sum under the wave but I couldn`t solve it. Can someone help me?

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回答（2 个）

Star Strider 2017-1-18

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Use the vpa function:

syms wn t 
wn = sym(1);
f = (wn.*t).*exp(1i.*2.*pi.*t);
f_int = int(f, t, 0, 0.3)
f_int_num = vpa(f_int)
f_int = 
- 1/(4*pi^2) - (((pi*3i)/5 - 1)*(1/4 + (2^(1/2)*(5^(1/2) + 5)^(1/2)*1i)/4 - 5^(1/2)/4))/(4*pi^2)
f_int_num = 
0.012251815898938149373515863015179 + 0.038845017631697804582142824751429i

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Star Strider 2017-1-18

My pleasure.

Please let me know.

Star Strider 2017-1-18

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To define ‘wn(t)’ as uniformly equal to 1 definitely changes the result:

syms wn t u_lim wn(t)
wn(t) = sym(1);
f = wn*exp(1i*2.*pi*t);
upper_limit = vpasolve(abs(int(f, t, 0, u_lim)) == 0.3, u_lim)
abs_upper_limit = abs(upper_limit)
upper_limit =
61.162453359143770665259861917249 - 0.1257699093208763500021861131513i
abs_upper_limit =
61.162582670939654308556507428042

Is the rest correct? Are you solving for the upper limit of integration that will make the integral equal to 0.3? If so, this works.

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David Goodmanson 2017-1-18

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编辑：David Goodmanson 2017-1-18

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Hello Diana, symbolic variables are a great thing, but if you are looking for a numerical result and are happy with 15 or so sigfigs, it isn't like they have to be invoked. You can just do

ff = @(t,wn) (wn.*t).*exp(1i.*2.*pi.*t)  % or you could define this in an mfile
integral(@(t) ff(t,1),0,.3)              % pass in wn =1
format long
ans =  0.012251815898938 + 0.038845017631698i

Now that symbolic variables are much better integrated into Matlab, sometimes I wonder if they are getting overused.

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Pilar Jiménez 2017-1-18

Yes, I will share with him to check it. Thanks for you support.

Pilar Jiménez 2017-1-18

I checked with my professor and I have a mistake, the function is incorrect. Wn is not multiplied by t, it is in function of t. I mean, the correct expression is wn(t) not wn.*t, do you know how I can register that in the code?

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