The answer is Humongous!
I = ∫ 0 ∞ x sin ( x ) e r f i ( x ) e − a x d x If I can be expressed as (for a > 1 ): I = ln g ( a ) f ( a ) Submit the value of ⌊ 1 0 0 0 0 0 ∗ f ( π ) ∗ g ( e ) ⌋ . Note that e r f i ( x ) is the Imaginary Error Function .
The answer is 908127.
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1 solution
This can be found on MSE . The result is proved for a = 2 and stated for general a > 1 ; the general proof is a simple extension of the specific case.
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Woah! Please tell me you are an active user in MSE as well. I would love to read your questions/comments/answers as well because they are incredibly insightful, as always.
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No, one site is enough. I check MSE out to see what's there, though.
Got a proof?
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Wait a little more! I'm expecting Mr hennings to give some insight!
The integral is: ∫ 0 ∞ x sin ( x ) e r f i ( x ) e − a x d x = ln a 2 − 2 a + 2 − 2 a 2 − 2 a + 2 − a + 1 + 1 a 2 − 2 a + 2 + 2 a 2 − 2 a + 2 − a + 1 + 1