Wait, That's Not Integrable!

Calculus Level 2

∫ 0 1 sin ⁡ x 2 d x \int_{0}^{1}\sin x^2 \, dx

Which of the following series can be expressed as the value of the integral above?

Hint: Take its Maclaurin series.

∑ n = 0 ∞ ( − 1 ) n ( 4 n + 3 ) ( 2 n + 1 ) ! \sum_{n = 0}^{\infty} \frac{(-1)^n}{(4n + 3)(2n + 1)!} ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) ! \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n + 1)!} ∑ n = 0 ∞ ( − 1 ) n ( 4 n + 2 ) ( 2 n + 1 ) ! \sum_{n = 0}^{\infty} \frac{(-1)^n}{(4n + 2)(2n + 1)!} ∑ n = 0 ∞ ( − 1 ) n ( 4 n + 1 ) ( 2 n + 1 ) ! \sum_{n = 0}^{\infty} \frac{(-1)^n}{(4n + 1)(2n + 1)!}

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Andrew Ellinor by Andrew Ellinor , 33, USA

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1 solution

∫ 0 1 sin ⁡ ( x 2 ) d x = ∫ 0 1 ∑ n = 0 ∞ ( − 1 ) n ( x 2 ) 2 n + 1 ( 2 n + 1 ) ! d x = ∫ 0 1 ∑ n = 0 ∞ ( − 1 ) n x 4 n + 2 ( 2 n + 1 ) ! d x = [ ∑ n = 0 ∞ ( − 1 ) n x 4 n + 3 ( 4 n + 3 ) ( 2 n + 1 ) ! ] 0 1 = ∑ n = 0 ∞ ( − 1 ) n ( 4 n + 3 ) ( 2 n + 1 ) ! \begin{aligned} \int_0^1 \sin{(x^2)} dx & = \int_0^1 \sum_{n=0}^\infty \frac {(-1)^n(x^2)^{2n+1}} {(2n+1)!} dx \\ & = \int_0^1 \sum_{n=0}^\infty \frac {(-1)^nx^{4n+2}} {(2n+1)!} dx \\ & = \left[ \sum_{n=0} ^\infty \frac {(-1)^n x^{4n+3}} {(4n+3)(2n+1)!} \right]_0^1 \\ & = \boxed{ \displaystyle \sum_{n=0} ^\infty \frac {(-1)^n} {(4n+3)(2n+1)!}} \end{aligned}

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süper COMRADE

Soner Karaca - 5 years, 3 months ago

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Thanks comrade

Chew-Seong Cheong - 1 year, 7 months ago

In the example of integral of sinx/x, there is an error. The integral of the series was not taken before evaluating.

Charles Mooney - 1 year, 7 months ago

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