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janhardo

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@acer Thanks  ......

janhardo 700
December 24 2021
0 0

@acer  Thanks

> 

restart;with(plots):
> 

A := Int(1/(b^2 + zz^2), zz = 0 .. z);

Int(1/(b^2+zz^2), zz = 0 .. z)

 (1)
> 

value(A);

arctan(z/b)/b

 (2)
> 

convert(value(A),ln);

((1/2)*I)*(ln(1-I*z/b)-ln(1+I*z/b))/b

 (3)
> 

combine(expand(I*convert(value(A),ln)), symbolic)/I;

-((1/2)*I)*ln(-(I*z+b)/(I*z-b))/b

 (4)

 

A little manipulation can grind the above into a form with the target leading sign, etc.

 

> 

evalindets(%, specfunc(ln),
           u->ln(expand(sign(numer(op(u)))*numer(op(u)))
                 /expand(sign(numer(op(u)))*denom(op(u)))));

-((1/2)*I)*ln((I*z+b)/(-I*z+b))/b

 (5)
> 

ans := sort(%, order=plex(b,z));

-((1/2)*I)*ln((b+I*z)/(b-I*z))/b

 (6)
> 

simplify( combine( convert( value(A) - ans, ln ), symbolic) );

0

 (7)
> 

 

> 

 

> 

 

> 

plot3d([Re,Im](arctan(z/b)/b-ans),b=-4..4,z=-4..4,
       color=[red,blue],grid=[100,100],view=-1e-14..1e-14);

 
> 

restart;
> 

A := Int(1/(b^2 + z^2), z = 0 .. z);

Int(1/(b^2+z^2), z = 0 .. z)

 (8)
> 

value(A);

arctan(z/b)/b

 (9)
> 

A1:=convert(value(A),ln);

((1/2)*I)*(ln(1-I*z/b)-ln(1+I*z/b))/b

 (10)
> 

A2:=expand(A1):%=combine(%);

((1/2)*I)*ln(1-I*z/b)/b-((1/2)*I)*ln(1+I*z/b)/b = ((1/2)*I)*(ln(1-I*z/b)-ln(1+I*z/b))/b

 (11)
> 

A3:=simplify(A2,symbolic);

((1/2)*I)*(ln(-I*z+b)-ln(I*z+b))/b

 (12)
> 

A4:=denom(A3);

2*b

 (13)
> 

A5:=numer(A3);

I*(ln(-I*z+b)-ln(I*z+b))

 (14)
> 

simplify(A5,symbolic);

I*(ln(-I*z+b)-ln(I*z+b))

 (15)
> 

I*ln(-z*I + b) - ln(z*I + b)= I*ln ((-z*I + b)/(z*I + b));

I*ln(-I*z+b)-ln(I*z+b) = I*ln((-I*z+b)/(I*z+b))

 (16)
> 

 

> 

 

And so on.. If some  of the high level commands are advanced, then a step back
If i can make (16)  from I*ln(-z*I + b) - ln(z*I + b)
a fraction?

> 

 

Download handmatig_betrekking_logaritmisch_en_goniometrisch.mw

@acerThanksSo your claim about my Maple ...

janhardo 700
December 24 2021
0 0

@acer

Thanks

So your claim about my Maple result being "almost correct" except that I "is still there" is not right. 

---------------------------

I overlooked the  i  ..it is  in there  , you are right. ( comes that i am so used at the normal variables naming) 
I think you have checked that your answer with simplify if it was correct ?
Cannot recognize yet the book answer with your answer unfortanely if both are the same 

@vv  Informative website for me ge...

janhardo 700
December 24 2021
0 0

@vv 

Informative website for me getting more insight in the complex analysis

Complex Analysis (complex-analysis.com)

@acer This answer you got has ...

janhardo 700
December 24 2021
0 0

@acer

This answer you got has  minus sign and a I inside , but  the bookanswer has no  - and a I next to  ln  
It proves how you must be careful to type the right expression, but is all a little bit complicated to get this expression not in Maple input here as post 

 

 -((1/2)*I)*ln((b+I*z)/(b-I*z))/b

@janhardo   ....

janhardo 700
December 24 2021
0 0

@janhardo 

> 

 

> 

         
J:= Int(exp(t*x), x);
J1:= subs(t= alpha+beta*I, J);
evalc(J1);
value(%);
simplify(convert(diff(%, x), exp), {a+b*I= t});

Int(exp(t*x), x)

  

Int(exp((alpha+I*beta)*x), x)

  

Int(exp(x*alpha)*cos(x*beta), x)+I*(Int(exp(x*alpha)*sin(x*beta), x))

  

alpha*exp(x*alpha)*cos(x*beta)/(alpha^2+beta^2)+beta*exp(x*alpha)*sin(x*beta)/(alpha^2+beta^2)+I*(-beta*exp(x*alpha)*cos(x*beta)/(alpha^2+beta^2)+alpha*exp(x*alpha)*sin(x*beta)/(alpha^2+beta^2))

  

exp((alpha+I*beta)*x)

 (1)
> 

#==========================================
> 

restart;         
J:= Int(exp(t*x), x);
J1:= subs((SUB:= t= alpha+beta*I), J);
J2:= evalc([Re,Im](J1));
J3:= J2 =~ simplify(value(J2));
print~(J3):
 
#reversion to original integrand:
simplify(
    convert(diff(evalc(Complex(rhs~(J3)[])), x), exp),
    {rhs=lhs}(SUB)
);

Int(exp(t*x), x)

  

Int(exp((alpha+I*beta)*x), x)

  

[Int(exp(x*alpha)*cos(x*beta), x), Int(exp(x*alpha)*sin(x*beta), x)]

  

[Int(exp(x*alpha)*cos(x*beta), x) = exp(x*alpha)*(beta*sin(x*beta)+cos(x*beta)*alpha)/(alpha^2+beta^2), Int(exp(x*alpha)*sin(x*beta), x) = exp(x*alpha)*(sin(x*beta)*alpha-beta*cos(x*beta))/(alpha^2+beta^2)]

  

Int(exp(x*alpha)*cos(x*beta), x) = exp(x*alpha)*(beta*sin(x*beta)+cos(x*beta)*alpha)/(alpha^2+beta^2)

  

Int(exp(x*alpha)*sin(x*beta), x) = exp(x*alpha)*(sin(x*beta)*alpha-beta*cos(x*beta))/(alpha^2+beta^2)

  

exp(t*x)

 (2)
> 

#================================================
> 

 

The difference in code starts with  evalc and the use of  J,J1,J2 and J3 makes it easier to read

Its a difficult command : J2:= evalc([Re,Im](J1)); in order to get rid of the I symbol, and probably there is no need for?
I must study further this to see how it excactly works.

 

Can you explain in your own words how this second code works ?       

Download complex_integration-tom_apostel_example_-bernouille.mw

@acer ThanksIt was tricky, because ...

janhardo 700
December 24 2021
0 0

@acer 

Thanks

Your code is also a difficult one to understand quick , so perhaps you can explain this in your own words too for me to get a idea for me.

It was tricky, because this  expression written (see hereunder)i s not the same in Holland or America/England? 
Log and ln are the same in America ( and in the book example too probably) and not in Holland
Log and ln are different. A = -(int(1/t, t = 1 .. t))/(2*bi) and -(int(1/t, t = 1 .. t))/(2*bi) = log[10](t)/(2*bi) and log[10](t)/(2*bi) = log[10]((b*i+z)/(b*i-z))/(2*bi)

So log written in the formula must read as ln then.
Your answer is almost correct , its only the I (imaginairy unit) is still there

In a another example of calculating a complex integral is used to remove the I   ....see hereunder

J:= Int(exp(t*x), x);
J1:= subs((SUB:= t= alpha+beta*I), J);
J2:= evalc([Re,Im](J1));

@Carl Love ThanksThere will be a re...

janhardo 700
December 24 2021
0 0

@Carl Love 

Thanks

There will be a reason why maple uses  log and ln for the same thing ..yes use in English math books ? 
In written math books here in Holland there is a distinction between the two: log and ln and they differ in meaning
Probably not in English math books : log and ln are the same then.

@Carl Love ThanksI noticed the diff...

janhardo 700
December 24 2021
0 0

@Carl Love 

Thanks

I noticed the difference between input and output coloring, so its a mistake 
This second code piece construct the two wanted integrals from 

 J := Int(exp(t*x), x)

Amazing second code piece ,but it needs some clarifcation compared with the first code piece. 

@Carl Love I used 2d input expressi...

janhardo 700
December 24 2021
0 0

@Carl Love 

I used 2d input expression palette and was writing , so it is not executable math then.
Was not further thinking on it , but it is possible to use the log and  ln for the same logaritme with  base e  : 

log[10](100);    can be used also in worksheet .
                             2
The book example use log as notation ( i assume with base 10 ) 

A  2 d input example                         

log[10](100);    
               2
ln(100.0);

             4.605170186
 

@janhardo Complex integration is so...

janhardo 700
December 23 2021
0 0

@janhardo 

Complex integration is somewhere started and ...

-------------------------------------------------------------------------------------------------info------------------------

So far, we’ve seen how to evaluate integrals of simple functions of a complex variable—that were defi ned in terms of a single real parameter we called t. Now it’s time to generalize and consider a more general case, where we just say we’re integrating a function of a complex variable f (z ), where ∈ C. This can be done using a technique called contour integration. The reason integrals of complex functions are done the way they are is that while an integral of a real-valued function is defi ned on an interval of the line, an integral  of a complex-valued function is defi ned on a curve in the complex plane.

@acer ThanksYes, there is leading m...

janhardo 700
December 23 2021
0 0

@acer 

Thanks

Yes, there is leading minus sign ( as you mean A= - ....) , but forget a second one after the first one  A = -.. = - ...= 

A = -(int(1/t, t = 1 .. t))/(2*bi) and -(int(1/t, t = 1 .. t))/(2*bi) = log[10](t)/(2*bi) and log[10](t)/(2*bi) = log[10]((b*i+z)/(b*i-z))/(2*bi)

@vv ThanksThis example is from Joha...

janhardo 700
December 23 2021
0 0

@vv 

Thanks

This example is from Johann Bernoulli [1702] who started with  A = .....
Interesting to know when the  "complex path integral". is invented ?

Do you think that Johann Bernoulli [1702] has this also in mind what you explained ?

@janhardo Don't get the right i...

janhardo 700
December 23 2021
0 0

@janhardo 
Don't get the right integral see , why not 

> 

restart;
> 

J:= Int(exp(t*x), x);
#J1:= subs(t= a+b*I, J);# heeft Carl gebruikt
J1:= subs(t= alpha+beta*I, J);
evalc(J1);
value(%);
simplify(convert(diff(%, x), exp), {a+b*I= t});
                            exp(x t)

Int(exp(t*x), x)

  

Int(exp((alpha+I*beta)*x), x)

  

Int(exp(x*alpha)*cos(x*beta), x)+I*(Int(exp(x*alpha)*sin(x*beta), x))

  

alpha*exp(x*alpha)*cos(x*beta)/(alpha^2+beta^2)+beta*exp(x*alpha)*sin(x*beta)/(alpha^2+beta^2)+I*(-beta*exp(x*alpha)*cos(x*beta)/(alpha^2+beta^2)+alpha*exp(x*alpha)*sin(x*beta)/(alpha^2+beta^2))

  

exp((alpha+I*beta)*x)

  

Error, missing operator or `;`

  
> 

 

It is theorem about complex exponentials|, in this case with exp(1)as base

If f(x)=exp(t*x)for all real x and a fixed complex t , then f '(x) =  texp(1)"^(t x) "

 

If you take the integral : int(e^(t*x), x) = e^(t*x)/t(1)

When t ≠ 0, if we let t = α + i*betaand t = α +β and equate the real and imaginairy parts of equation (1)  , we obtain the integration formulas

 

"∫e^(alpha x)cos beta x ⅆx = " exp(alpha*x)*(alpha*cos(beta*x)+beta*sin(beta*x))/(alpha^2+beta^2)

int(e^(alpha*x)*sin*beta*x, x) = 2*exp(alpha*x)*(alpha*sin(beta*x)+beta*cos(beta*x))/(alpha^2+beta^2)

wich are valid if α and β are not both zero

 

 

Consider the differential equation :
`y"`+a*(diff(y(x), x))+b*y(x) = 0
The real and imaginairy parts of the function f(x)=exp(t*x)are solutions of this DE
Now as exercise i can solve this DE to check this
This exercise was for me how to use/find commands for complex variables

> 

#alpha*exp(x*alpha)*cos(x*beta)/(alpha^2 + beta^2) + beta*exp(x*alpha)*sin(x*beta)/#(alpha^2 + beta^2);
> 

 

> 

#normal(%);
> 

#simplify(%);
> 

#a:=-beta*exp(x*alpha)*cos(x*beta)/(alpha^2 + beta^2) + alpha*exp(x*alpha)*sin(x*#beta)/(alpha^2 + beta^2);
> 

#simplify(a);
> 

restart;
> 

Int(exp(x*alpha)*cos(x*beta), x)= alpha*exp(x*alpha)*cos(x*beta)/(alpha^2 + beta^2) + beta*exp(x*alpha)*sin(x*beta)/(alpha^2 + beta^2);

Int(exp(x*alpha)*cos(x*beta), x) = alpha*exp(x*alpha)*cos(x*beta)/(alpha^2+beta^2)+beta*exp(x*alpha)*sin(x*beta)/(alpha^2+beta^2)

 (1)
> 

simplify(%);

Int(exp(x*alpha)*cos(x*beta), x) = exp(x*alpha)*(cos(x*beta)*alpha+sin(x*beta)*beta)/(alpha^2+beta^2)

 (2)
> 

Int(exp(x*alpha)*sin(x*beta), x)= -beta*exp(x*alpha)*cos(x*beta)/(alpha^2 + beta^2) + alpha*exp(x*alpha)*sin(x*beta)/(alpha^2 + beta^2);

Int(exp(x*alpha)*sin(x*beta), x) = -beta*exp(x*alpha)*cos(x*beta)/(alpha^2+beta^2)+alpha*exp(x*alpha)*sin(x*beta)/(alpha^2+beta^2)

 (3)
> 

simplify(%);

Int(exp(x*alpha)*sin(x*beta), x) = -exp(x*alpha)*(cos(x*beta)*beta-sin(x*beta)*alpha)/(alpha^2+beta^2)

 (4)
> 

Int(exp(x*alpha)*sin(x*beta), x)*I= (-beta*exp(x*alpha)*cos(x*beta)/(alpha^2 + beta^2) + alpha*exp(x*alpha)*sin(x*beta)/(alpha^2 + beta^2))*I;

(Int(exp(x*alpha)*sin(x*beta), x))*I = (-beta*exp(x*alpha)*cos(x*beta)/(alpha^2+beta^2)+alpha*exp(x*alpha)*sin(x*beta)/(alpha^2+beta^2))*I

 (5)
> 

 

Download complexe_getallen_post_3_integral_(1).mw

@Carl Love Thanks  ......

janhardo 700
December 23 2021
0 0

@Carl Love 

Thanks
Yes, when i evaluate J1 integral i do get two integrals with their values
With some rearrangement i should get them in the wanted form?

> 

restart;
> 

J:= Int(exp(t*x), x);
J1:= subs(t= a+b*I, J);
evalc(J1);
value(%);
simplify(convert(diff(%, x), exp), {a+b*I= t});
                            exp(x t)

Int(exp(t*x), x)

  

Int(exp((a+I*b)*x), x)

  

Int(exp(x*a)*cos(x*b), x)+I*(Int(exp(x*a)*sin(x*b), x))

  

a*exp(x*a)*cos(x*b)/(a^2+b^2)+b*exp(x*a)*sin(x*b)/(a^2+b^2)+I*(-b*exp(x*a)*cos(x*b)/(a^2+b^2)+a*exp(x*a)*sin(x*b)/(a^2+b^2))

  

exp(t*x)

  

Error, missing operator or `;`

  
> 

 

It is theorem about complex exponentials|, in this case with exp(1)as base

If f(x)=exp(t*x)for all real x and a fixed complex t , then f '(x) =  texp(1)"^(t x) "

 

If you take the integral : int(e^(t*x), x) = e^(t*x)/t(1)

When t ≠ 0, if we let t = α + i*betaand t = α +β and equate the real and imaginairy parts of equation (1)  , we obtain the integration formulas

 

"∫e^(alpha x)cos beta x ⅆx = " exp(alpha*x)*(sin(x*beta)*beta+cos(x*beta)*alpha)/(alpha^2+beta^2)

"∫e^(alpha x)sin beta x ⅆx = "??

wich are valid if α and β are not both zero

 

 

Consider the differential equation :
`y"`+a*(diff(y(x), x))+b*y(x) = 0
The real and imaginairy parts of the function f(x)=exp(t*x)defined on whole x -axis, are solutions of this DE
Now as exercise i can solve this DE to check this
This exercise was for me how to use/find commands for complex variables

Download complexe_getallen_post_3_integral.mw

@acer ThanksYes, i know already ear...

janhardo 700
December 22 2021
0 0

@acer 
Thanks

Yes, i know already earlier (realized)  that it make no sense the diff(%) command here.
That was the next step for me to figure out .( there was a x in it and other variables were symbolic )
I got the same result as in textbook with maple for this integral.
Unfortanely its still difficult to see the relation with the start integral and this integral ? 

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