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- Thread starter Mr Davis 97
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- #1

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- #2

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Can you not change variables? Let ##u=tx##, etc. etc.

- #3

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Okay, I see how that could work... But I don't know the sign of t, right? As such I can't tell whether the upper bound on the integral goes to ##+ \infty## or ##- \infty##Can you not change variables? Let ##u=tx##, etc. etc.

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- #5

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Is there a reason for why you don't deal with the original integral via calculus of residues?

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