Is the Integral of an Odd Function with a Non-Centered Even Exponential Zero?

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SUMMARY

The integral of the function \(\int^{\infty}_{-\infty} ie^{-(x-x_{0})^{2}} \sin(vx) dx\) is analyzed to determine if it equals zero. The sine function is confirmed as odd, while the exponential function \(e^{-(x-x_{0})^{2}}\) is not centered around the origin, leading to the conclusion that the integral does not necessarily equal zero. The discussion suggests using the trigonometric identity for sine to further explore the integral's behavior.

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atlantic
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Homework Statement



I have the integral:
\int^{\infty}_{-\infty}ie^{-(x-x_{0})^{2}}sin(vx) dx

where x_0 and v are real constants.

The sine function is odd. But what about the exponential? If it's even, then the integral is zero, but the exponential is not centred around origo. Will the integral still be zero as long as the exponential is not an odd function?
 
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hi atlantic! :smile:

no

try using sin(vx) = sin(v(x - xo))cos(vxo) + cos(v(x - xo))sin(vxo) :wink:
 
The first thing I would do is to let x=\pm a and compare the two, what would you expect to happen if the function was odd?
 

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