Integrate sin(u)/u with respect to u (related to the Sine integral Si(x)?

using integration by parts,

Int [(sin u)/u] du = sin u * ln u - Int cos u * ln u du

Int cos u ln u du = sin u ln u - Int (sin u)/u du

So

Int [(sin u)/u] du = sin u * ln u - sin u ln u + Int (sin u)/u du

or 0 = 0.

So this method does not work...

By the way, the Sine integral Si(x) is defined to be

Si(x) = Int[0 to x] [(sin u)/u] du

1 Answer

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  • Dr D
    Lv 7
    1 decade ago
    Favorite Answer

    This result is not achieved using integration by parts.

    First of all Si(x) is defined as Int[0 to x] [(sin u)/u] du as you wrote.

    So essentially ∫ sin(u)/u du = Si(u) + constant

    Secondly, this result can be derived using the Fourier integral. It can be shown that

    ∫{0,∞} sin(u)/u du = π/2 [Dirichlet integral]

    Then the sine integral, is just defined to be the definite integral from 0 to x, such that

    Si(∞) = π/2, Si(0) = 0, etc

    You can start with f(x) = 1 if |x| < 1, 0 otherwise.

    Then represent f(x) as a Fourier integral:

    f(x) = 2/π ∫{0,∞} sin(u) cos(ux) / u du

    Now set x = 0

    f(0) = 2/π ∫{0,∞} sin(u) / u du

    And we know from the definition that f(0) = 1

    So ∫{0,∞} sin(u) / u du = π/2

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