- #1

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Why we suppose that all singularities are left from line [tex]Re(s)=c[/tex]?

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- Thread starter matematikuvol
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- #1

- 192

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Why we suppose that all singularities are left from line [tex]Re(s)=c[/tex]?

- #2

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- #3

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Why we suppose that all singularities are left from line [tex]Re(s)=c[/tex]?

We place the path of integration (by design and not by necessity), to the right of all singularities so that the value of the integral is independent of the specific value of c. You can show that by evaluating the integral over a square contour which goes around to the right and since there are no singularities there, the value of the integral is zero. Now due to the restrictions placed on functions which have laplace transforms (those of exponential order), the horizontal legs on the top and bottom of that contour can be shown to be zero which means the sum of the two vertical legs are zero which means they are equal to one another when both are going in the same direction which means the integral is independent of the value of c.

- #4

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My second question is:

When I defined Laplase transform like

[tex]F(s)=\int^{\infty}_0f(t)e^{-st}dt[/tex]

I say that [tex]Re(s)>0[/tex] because of convergence. So [tex]c[/tex] must be positive. Right? Is there some other condition?

- #5

- 192

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http://www.solitaryroad.com/c916.html

Why integral [tex]\int_{C_1}=0[/tex] when [tex]R\rightarrow \infty[/tex]? I have

[tex]\lim_{R\to \infty}\int_{C_1}F(s)e^{st}dp=0[/tex]

If I understand [tex]R\rightarrow \infty[/tex] is equivalent with [tex]\lim_{Re(s)\to \infty}[/tex].

[tex]\lim_{Re(s)\to \infty}F(s)=0[/tex]

and

[tex]\lim_{Re(s)\to \infty}e^{pt}=\infty[/tex]

why then

[tex]\lim_{R\to \infty}\int_{C_1}F(s)e^{st}dp=0[/tex]?

- #6

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Can someone help me and answer these questions?

- #7

- 192

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Here is picture of my question. Can you give me detail explanation?

Can you give me a explanation why all singularities are left from ##Re(s)=c##? And why we integrate over the line ##(c-i\infty,c+i\infty)##?

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