I have to see if F ( t ) = ∫ 0 ∞ sin ⁡ ( t x ) e − x 2 t x 3 / 2 d xis well defined when t &gt; 0 and when t &lt; 0.For it, I have to see if | F ( t ) | &lt; ∞. I have done the following | F ( t ) | = | ∫ 0 ∞ sin ⁡ ( t x ) e − x 2 t x 3 / 2 d x | ≤ ∫ 0 ∞ | sin ⁡ ( t x ) e − x 2 t x 3 / 2 | d x = ∫ 0 ∞ | sin ⁡ ( t x ) | e − x 2 t x 3 / 2 d x = ∫ 0 1 | sin ⁡ ( t x ) | e − x 2 t x 3 / 2 d x + ∫ 1 ∞ | sin ⁡ ( t x ) | e − x 2 t x 3 / 2 d x ≤ ∫ 0 1 t e − x 2 t x 1 / 2 d x + ∫ 1 ∞ e − x 2 t x 3 / 2 d x 1. When t &gt; 0: The first integral ∫ 0 1 t e − x 2 t x 1 / 2 d x ≤ t ∫ 0 1 1 x 1 / 2 d x = 2 t &lt; ∞(I can say that it is finite, right?). But the second one?2. And when t &lt; 0, what can we say?I don't know how to bound them.

Question

I have to see if  F  (  t  )  =      ∫    0          ∞                  sin      ⁡      (      t      x      )              e                  −                      x            2                    t                            x              3                  /                2              d  xis well defined when   t  &amp;gt;  0 and when   t  &amp;lt;  0.For it, I have to see if       |    F  (  t  )      |    &amp;lt;  ∞. I have done the following                    |        F        (        t        )        |                            =                              |                                    ∫          0                      ∞                                                sin            ⁡            (            t            x            )                          e                              −                                  x                  2                                t                                                          x                          3                              /                            2                                      d        x                              |                                                            ≤                  ∫          0                      ∞                                                |                                                sin            ⁡            (            t            x            )                          e                              −                                  x                  2                                t                                                          x                          3                              /                            2                                                            |                          d        x                                          =                  ∫          0                      ∞                                                |            sin            ⁡            (            t            x            )            |                          e                              −                                  x                  2                                t                                                          x                          3                              /                            2                                      d        x                                          =                  ∫          0                      1                                                |            sin            ⁡            (            t            x            )            |                          e                              −                                  x                  2                                t                                                          x                          3                              /                            2                                      d        x        +                  ∫          1                      ∞                                                |            sin            ⁡            (            t            x            )            |                          e                              −                                  x                  2                                t                                                          x                          3                              /                            2                                      d        x                                          ≤                  ∫          0                      1                                                t                          e                              −                                  x                  2                                t                                                          x                          1                              /                            2                                      d        x        +                  ∫          1                      ∞                                                e                          −                              x                2                            t                                            x                          3                              /                            2                                      d        x            1. When   t  &amp;gt;  0: The first integral      ∫    0          1                  t              e                  −                      x            2                    t                            x              1                  /                2              d  x  ≤  t      ∫    0          1            1          x              1                  /                2              d  x  =  2  t  &amp;lt;  ∞(I can say that it is finite, right?). But the second one?2. And when   t  &amp;lt;  0, what can we say?I don&#039;t know how to bound them.

Blaze Frank · Accepted Answer

ℜ  (  t  )  &amp;gt;  0 is required.Now, using what you wrote is  |  F  (  t  )  |  ≤      ∫    0          1                  t              e                  −                      x            2                    t                            x              1                  /                2              d  x  +      ∫    1          ∞                  e              −                  x          2                t                    x              3                  /                2              d  x      I    1    =      ∫    0          1                  t              e                  −                      x            2                    t                            x              1                  /                2              d  x  =      1    2        t          3              /            4            (    Γ          (              1        4            )        −    Γ          (              1        4            ,      t      )        )  For the second one, one integration by parts leads to  J  =  ∫            e              −                  x          2                t                    x              3                  /                2              d  x  =            2              (                  t                      1                          /                        4                                    x                        Γ                  (                      3            4                    ,          t                      x            2                    )                −                  e                      −            t                          x              2                                      )                    x            I    2    =      ∫    1    ∞              e              −                  x          2                t                    x              3                  /                2              d  x  =  2      e          −      t        −  2      t          1              /            4        Γ      (          3      4        ,    t    )    |  F  (  t  )  |  ≤      1    2        t          3              /            4            (    Γ          (              1        4            )        −    Γ          (              1        4            ,      t      )        )    +  2      e          −      t        −  2      t          1              /            4        Γ      (          3      4        ,    t    )

I have to see if F ( t ) = <msubsup> ∫ 0 <mrow

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