INTEGRATE it

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prove \mathbf{}\int_{0}^{infinity}{sinx.dx}=1

#Mathematics #Calculus

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8 Answers

1

" ____________ ·2010-02-23 06:27:24

integral does not converge

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1

Manmay kumar Mohanty ·2010-02-23 06:35:17

ya not integrable.....

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39

Dr.House ·2010-02-23 06:54:08

yes it fails to exist...

this is actually grandi`s series in an integral form

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39

Dr.House ·2010-02-23 06:54:35

and grandi`s series is nothing but 1-1+1-1+1.........................

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1

b_k_dubey ·2010-02-23 09:19:26

I=\int_{0}^{\infty }{e^{-ax}sinx}\: dx = -\left[ \frac{e^{-ax}(asinx+cosx)}{a^2+1}\right]_{0}^{\infty }

I=\frac{1}{a^2+1}

required integral is obtained by putting a=0 which gives I = 1

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39

Dr.House ·2010-02-23 09:32:57

sir itw a sfor students... kindly hide or remove your post...

it would help other students to think

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66

kaymant ·2010-02-23 09:34:34

okay...i removed my post b555

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Lokesh Verma ·2010-02-23 13:00:00

rahul.. where did you get this questoin from?

and please try to understand what bhargav is saying

It may help.

also try to find the flaw in bipin's post!

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Your Answer

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