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Use the weighted mean value theorem to establish an integral formula

Prove that for any positive integer n,

    \[ \int_{\sqrt{n \pi}}^{\sqrt{(n+1)\pi}} \sin (t^2) \, dt = \frac{(-1)^n}{c}, \qquad \text{for } \sqrt{n \pi} \leq c \leq \sqrt{(n+1)\pi}. \]

Recall the weighted mean value theorem:

For functions f and g continuous on [a,b], if g never changes sign in [a,b] then there exists c \in [a,b] such that

    \[ \int_a^b f(x) g(x) \, dx = f(c) \int_a^b g(x) \, dx. \]

Proof. In order to apply the weighted mean value theorem (Theorem 3.16 of Apostol) we need to identify the functions f and g of the theorem. So, let

    \[ f(t) = \frac{1}{t}, \qquad g(t) = t \sin (t^2). \]


    \[ \int_{\sqrt{n \pi}}^{\sqrt{(n+1)\pi}} \sin (t^2) \, dt = \int_{\sqrt{n \pi}}^{\sqrt{(n+1) \pi}} \frac{t \sin (t^2)}{t} \, dt = \int_{\sqrt{n \pi}}^{\sqrt{(n+1)\pi}} f(t) g(t) \, dt. \]

Furthermore, for a positive integer n, g(t) = t \sin (t^2) does not change sign on the interval [\sqrt{n \pi}, \sqrt{(n+1)\pi}] (since \sin(t^2) and t do not change sign on this interval). So we may apply the theorem to conclude there exists c \in [\sqrt{n \pi}, \sqrt{(n+1)\pi} ] such that

    \[ \int_{\sqrt{n \pi}}^{(n+1)\pi} f(t) g(t) \, dt = f(c) \int_{\sqrt{n \pi}}^{\sqrt{(n+1) \pi}} g(t) \, dt = \frac{1}{c} \int_{\sqrt{n \pi}}^{\sqrt{(n+1) \pi}} t \sin (t^2) \, dt. \]

Now we need to evaluate this integral. (Here I’m going to cheat a little. In the book we do not yet have techniques available to evaluate this integral. We’ll develop the Fundamental Theorem of Calculus, etc, in Chapter 5 that will allow us to do this. If you have a way to do this without directly evaluating an integral that we don’t yet know how to evaluate please do leave a comment with your solution.)

    \[ \int_{\sqrt{n \pi}}^{\sqrt{(n+1) \pi}} t \sin(t^2) \, dt = -\frac{1}{2} \int_{\sqrt{n \pi}}^{\sqrt{(n+1) \pi}} -2 t \sin(t^2) \, dt = \left. -\frac{1}{2} \cos(t^2) \right|_{\sqrt{n \pi}}^{\sqrt{(n+1) \pi}}  = (-1)^n. \]

Where the final equality follows since \cos \left( \sqrt{(n+1) \pi}\right) - \cos \left( \sqrt{n \pi} \right) = -2 if n is even and equals 2 if n is odd. Then multiplying that by the -\frac{1}{2}, we get (-1)^n.

Putting this together we then have,

    \[ \int_{\sqrt{n \pi}}^{\sqrt{(n+1) \pi}} \sin (t^2) \, dt = \frac{(-1)^n}{c} \qquad c \in \left[ \sqrt{n \pi}, \sqrt{(n+1) \pi} \right]. \qquad \blacksquare\]


  1. Ishan Bhat says:

    Your algorithm is correct but a simpler solution could be by using substitution and then by parts. The answer comes out to be a simple one!

  2. nu creation says:

    The preface to the 2nd edition says, “…the mean-value theorems and routine applications are introduced at an earlier stage.” I bet in the original edition, this problem was in a section
    after the fundamental theorem of calculus was introduced.

  3. zz says:

    After spending upwards 10 hours trying to solve this without invoking a fundamental theorem, I’m glad to have found I’m not missing anything obvious.

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