calculate g(k)=integral(-infinite to +infinite)​f(x)*(e^(-​alxl))*e^(​-ikx)

Assume f(x+1)=f(x),how to calculate g(k)=integral(-infinite to +infinite)f(x)*(e^(-alxl))*e^(-ikx) in matlab.

Then,input different a and k in matlab to try to find different integral values.

[Chunru]

Are you sure the problem is well-defined? f(x+1)=f(x) indicate f(x) is an arbitrary periodic function with period of 1.

[John D'Errico]

PLEASE STOP ASKING THIS SAME QUESTION. This is the 9'th time you have asked it.

You are now becoming a spammer on the site.

Answer 1

[Walter Roberson]

syms x a k
syms f(x)
assume(a>0)
f(x) = sin(2*pi*x) + cos(6*pi*x)

f(x) = 

inner = f(x)*exp(-a*abs(x))*exp(-1i*k*x)

inner = 

g(k) = int(inner, x, -inf, inf)

g(k) = 

Notice that the limit() portions involve sin(2*pi*x) as x goes to infinity. However, that is not defined, so the limit does not exist, so the integral is not defined.

What about constant f(x) ?

f(x) = 5

f(x) = 

5

inner = f(x)*exp(-a*abs(x))*exp(-1i*k*x)

inner = 

g(k) = int(inner, x, -inf, inf)

g(k) = 

expand(rewrite(g,'sincos'))

ans(k) = 

... and we have the same problem, we have sin() and cos() of something involving x as x goes to infinity.

Your exp(-1i*k*x) is a phase term, and for constant k, as x goes to infinity you get infinite phase, which does not have a limit . The only way that you can get a definite integral in your f(x)*exp(-a*abs(x))*exp(-1i*k*x) expression as x goes to infinity, is if f(x) has a limit of 0. But we know that f(x+1) = f(x), so as x approaches infinity, for the limit f(x) to be 0, limit f(x) would have to be 0, so f(x) would have to be 0.

[Chris Lin]

May i ask what sigma1,sigma2...means and why you assume?

[Walter Roberson]

The sigma are temporary variables automatically created by the LiveScript display routines, for the purpose of making the output more compact. Any one of them does not necessarily have physical or mathematical meaning, possibly just coincidence. The system just looks for parts of the expression that are the same and substitutes a variable for the common portion.

For example suppose that the original expression was exp(a*x) + (b+1)*x, and suppose that you substitute a=2, b=1, getting exp(2*x) + 2*x. The 2 inside the exp() and the 2 outside the exp have no relationship to each other, but the display system just notices that they are the same and might choose to write

exp(sigma1) + sigma1 
where sigma1 = 2*x

so the generated sigma variables do not necessarily have any deep meaning.

[Walter Roberson]

In the expression exp(-a*abs(x)) with x -inf to +inf you have several possibilities

• if a is a negative real value then -a would be a positive value. Multiply by the abs(x) that is never negative and the result is never negative. exp() as x went to inf or - inf would grow to infinity and so the integral wound probably diverge
• if a is complex valued with nonzero imaginary part then as abs(x) increases, the result is undefined, because the rewrite in terms of sin() cos() would have indefinite sign
• if a is positive then -a*abs(x) is never positive and as x increases, the exp() decay to 0, leaving hope for a converging integral.

I will have another look at the limit when I am less tired; perhaps the limit is defined after all.

[Chris Lin]

thanks！！

[Walter Roberson]

I found the missing piece: you need to assume that k is real.

syms x a k real
syms f(x)
assume(a>0)
f(x) = sin(2*pi*x) + cos(6*pi*x)

f(x) = 

inner = f(x)*exp(-a*abs(x))*exp(-1i*k*x)

inner = 

g(k) = int(inner, x, -inf, inf)

g(k) = 

[Chris Lin]

How to draw g(k) when a=0 and k=+-2pi or k=+-6pi in the matlab?

[Walter Roberson]

You cannot plot that.

syms x real
syms K integer
k = 2*pi*K;
syms a real
f(x) = sin(2*pi*x) + cos(6*pi*x)

f(x) = 

inner = f(x)*exp(-a*abs(x))*exp(-1i*k*x)

inner = 

g(K) = int(inner, x, -inf, inf)

g(K) = 

limit(g(1), a, 0, 'right')

ans = 

limit(g(-1), a, 0, 'right')

ans = 

limit(g(3), a, 0, 'right')

ans = 

∞

limit(g(-3), a, 0, 'right')

ans = 

∞

[Chris Lin]

what is the meaning of limit(g(1), a, 0, 'right') Thanks

[Walter Roberson]

Limit from right: https://en.wikipedia.org/wiki/One-sided_limit

syms x real
syms k real
syms a real
f(x) = sin(2*pi*x) + cos(6*pi*x)

f(x) = 

inner = f(x)*exp(-a*abs(x))*exp(-1i*k*x)

inner = 

g(k) = int(inner, x, -inf, inf)

g(k) = 

g2pi = g(2*pi)

g2pi = 

gm2pi = g(-2*pi)

gm2pi = 

g6pi = g(6*pi)

g6pi = 

gm6pi = g(-6*pi)

gm6pi = 

limit(g2pi, a, 0, 'right')

ans = 

limit(gm2pi, a, 0, 'right')

ans = 

limit(g6pi, a, 0, 'right')

ans = 

∞

limit(gm6pi, a, 0, 'right')

ans = 

∞

subs(g2pi, a, 0)

Error using symengine
Division by zero.

Error in sym/subs>mupadsubs (line 169)
G = mupadmex('symobj::fullsubs',F.s,X2,Y2);

Error in sym/subs (line 154)
G = mupadsubs(F,X,Y);

subs(gm2pi, a, 0)
subs(g6pi, a, 0)
subs(gm6pi, a, 0)

[Chris Lin]

What is senario if f(x)= exp(2* pi*x)

It appears the error.

[Walter Roberson]

When you make then you remove the exponential dampning of the exp(-1i*k*x) term, and as I explained before that exp(-1i*k*x) term for real-valued k is a term that is not defined, as it involves sin(infinity) and cos(infinity)

[Chris Lin]

How to solve this situation ?

[Walter Roberson]

The only solution in that case is f(x) = 0