∫ 0 ∞ sin x 5 d x = a 1 ( b − 1 ) Γ ( c 1 )
The equation above holds true for positive integers a , b , and c . Submit a b c .
Notation: Γ ( ⋅ ) denotes the gamma function .
Hint: Use e i x = cos x + i sin x .
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The substitution i y = x 5 needs to be justified by contour integration. It cannot simply be used.
You need the fact that the integrand has no poles in the first quadrant, and that it tends to zero as ∣ z ∣ → ∞ .
Nice Solution. I too did it in the same way.
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I think this question is overrated...Do you think so?
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I am not particular on the rate. I think the system will resolve it when more members try it.
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@Chew-Seong Cheong – Yeah, that's true...but for the present, I feel it is so...
@Chew-Seong Cheong Sir, can you please verify the answer of this question
If you make the substitution i y = x 5 , then the bounds of integration on y become y = 0 → y = − i ∞ . In other words, an integration along the positive real axis in x corresponds to an integration along the negative imaginary axis in y . Your solution does not reflect this.
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You are right. See comments by Mark Hennings above.
I have revised the solution. Thanks for the comments.
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I = ∫ 0 ∞ sin x 5 d x = ∫ 0 ∞ 2 i ( e − i x 5 − e i x 5 ) d x = 2 i ∫ 0 ∞ e − i x 5 d x − 2 i ∫ 0 ∞ e i x 5 d x = 1 0 e − 5 2 π i ∫ 0 ∞ u − 5 4 e − u d u + 1 0 e 5 2 π i ∫ 0 ∞ u − 5 4 e − u d u = 5 cos 5 2 π Γ ( 5 1 ) = 2 0 1 ( 5 − 1 ) Γ ( 5 1 ) u = i x 5 ⟹ d u = 5 i x 4 d x u = − i x 5 ⟹ d u = − 5 i x 4 d x
⟹ a b c = 2 0 × 5 × 5 = 5 0 0