Simplify integral considering only the real part
$begingroup$
I happened to stumble on the following simplification of an integral:
$$ frac{1}{pi} int_{0}^{infty} dx e^{-ax} cdot cos(kx) = frac{1}{pi} Re left[ int_{0}^{infty} dx e^{x (ik - a)} right] $$
According to my reasoning: $ cos(kx) = frac{1}{2} left( e^{ikx} + e^{-ikx} right) $, so the term on the exponent on the right side comes somehow from the first term of the complex form of $cos(kx)$, but the second term gets cancelled, why?
integration complex-analysis analysis complex-integration
asked Jan 12 at 8:45
LeroyLeroy
1267
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add a comment |
$begingroup$
I happened to stumble on the following simplification of an integral:
$$ frac{1}{pi} int_{0}^{infty} dx e^{-ax} cdot cos(kx) = frac{1}{pi} Re left[ int_{0}^{infty} dx e^{x (ik - a)} right] $$
According to my reasoning: $ cos(kx) = frac{1}{2} left( e^{ikx} + e^{-ikx} right) $, so the term on the exponent on the right side comes somehow from the first term of the complex form of $cos(kx)$, but the second term gets cancelled, why?
integration complex-analysis analysis complex-integration
asked Jan 12 at 8:45
LeroyLeroy
1267
$endgroup$
add a comment |
$begingroup$
I happened to stumble on the following simplification of an integral:
$$ frac{1}{pi} int_{0}^{infty} dx e^{-ax} cdot cos(kx) = frac{1}{pi} Re left[ int_{0}^{infty} dx e^{x (ik - a)} right] $$
According to my reasoning: $ cos(kx) = frac{1}{2} left( e^{ikx} + e^{-ikx} right) $, so the term on the exponent on the right side comes somehow from the first term of the complex form of $cos(kx)$, but the second term gets cancelled, why?
integration complex-analysis analysis complex-integration
asked Jan 12 at 8:45
LeroyLeroy
1267
$endgroup$
I happened to stumble on the following simplification of an integral:
$$ frac{1}{pi} int_{0}^{infty} dx e^{-ax} cdot cos(kx) = frac{1}{pi} Re left[ int_{0}^{infty} dx e^{x (ik - a)} right] $$
According to my reasoning: $ cos(kx) = frac{1}{2} left( e^{ikx} + e^{-ikx} right) $, so the term on the exponent on the right side comes somehow from the first term of the complex form of $cos(kx)$, but the second term gets cancelled, why?
integration complex-analysis analysis complex-integration
integration complex-analysis analysis complex-integration
asked Jan 12 at 8:45
LeroyLeroy
1267
asked Jan 12 at 8:45
LeroyLeroy
1267
asked Jan 12 at 8:45
LeroyLeroy
1267
asked Jan 12 at 8:45
LeroyLeroy
1267
asked Jan 12 at 8:45
LeroyLeroy
1267
1267
add a comment |
add a comment |
1 Answer
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$begingroup$
Consider first $e^{ikx} = cos(kx) + i sin(kx)$ then you can easily deduce that
$$
Re(e^{ikx}) = cos(kx).
$$
answered Jan 12 at 8:59
WarreGWarreG
208110
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Consider first $e^{ikx} = cos(kx) + i sin(kx)$ then you can easily deduce that
$$
Re(e^{ikx}) = cos(kx).
$$
answered Jan 12 at 8:59
WarreGWarreG
208110
$endgroup$
add a comment |
$begingroup$
Consider first $e^{ikx} = cos(kx) + i sin(kx)$ then you can easily deduce that
$$
Re(e^{ikx}) = cos(kx).
$$
answered Jan 12 at 8:59
WarreGWarreG
208110
$endgroup$
add a comment |
$begingroup$
Consider first $e^{ikx} = cos(kx) + i sin(kx)$ then you can easily deduce that
$$
Re(e^{ikx}) = cos(kx).
$$
answered Jan 12 at 8:59
WarreGWarreG
208110
$endgroup$
Consider first $e^{ikx} = cos(kx) + i sin(kx)$ then you can easily deduce that
$$
Re(e^{ikx}) = cos(kx).
$$
answered Jan 12 at 8:59
WarreGWarreG
208110
answered Jan 12 at 8:59
WarreGWarreG
208110
answered Jan 12 at 8:59
WarreGWarreG
208110
answered Jan 12 at 8:59
WarreGWarreG
208110
208110
add a comment |
add a comment |
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