Integral from o to inf of sin(x/2)dx

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In summary, the conversation discusses the evaluation of the integral from 0 to infinity of sin(x/2)dx. The speaker first removes the infinity by taking the limit as a approaches infinity and then integrates the function. However, the limit does not exist and the integral is determined to be divergent.
  • #1
BoldKnight399
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Integral from o to inf of sin(x/2)dx

First thing I did was ditch the infinity so it became:
lim as a-->inf of integral from 0 to a of sin(x/2)dx

next I integrated it:
lim as a approaches infin of -cos(x/2) evaluated from o to a

then I plugged in:
lima->inf of -2cos(a/2)+2(cos0)

My problem is that I don't know if cos of inf diverges of converges. I know that it occelates, but what does that count as? Or, am I missing a step in here or just messed up somewhere along way? If anyone has any suggestions, I would greatly appreciate it.
 
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  • #2


You can't take the cosine of infinity. The limit of interest, lim a --> inf -2cos(a/2) does not exist, so your original integral diverges.
 
  • #3


ok Thank you!
 

FAQ: Integral from o to inf of sin(x/2)dx

What is the value of the integral from 0 to infinity of sin(x/2)dx?

The value of this integral is undefined, as it does not converge to a finite value. The integral oscillates between positive and negative values and extends to infinity, making it divergent.

How do you solve the integral from 0 to infinity of sin(x/2)dx?

This integral cannot be solved analytically using standard integration techniques. However, it can be approximated numerically using methods such as the trapezoidal rule or Simpson's rule.

What is the graph of the integral from 0 to infinity of sin(x/2)dx?

The graph of this integral will be a series of waves that oscillate between positive and negative values, with the amplitude increasing as x approaches infinity.

Why is the integral from 0 to infinity of sin(x/2)dx considered divergent?

This integral is considered divergent because it does not converge to a finite value. As x approaches infinity, the oscillations of the sine function become larger and larger, resulting in an infinite area under the curve.

Can the integral from 0 to infinity of sin(x/2)dx be evaluated using special functions?

No, this integral cannot be evaluated using special functions such as the error function or the gamma function. These functions are only defined for certain values of x and cannot be used to solve this divergent integral.

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