Integrate the product of sums!

Calculus Level 5

I = ∫ 0 ∞ ( x − x 3 2 + x 5 2 ⋅ 4 − x 7 2 ⋅ 4 ⋅ 6 + ⋯ ) ( 1 + x 2 2 2 + x 4 2 2 ⋅ 4 2 + x 6 2 2 ⋅ 4 2 ⋅ 6 2 + ⋯ ) d x \displaystyle I = \int_0^\infty {\left(x - \frac{x^3}{2} + \frac{x^5}{2 \cdot 4} - \frac{x^7}{2 \cdot 4 \cdot 6} + \cdots \right) \left(1 + \frac{x^2}{2^2} + \frac{x^4}{2^2 \cdot 4^2} + \frac{x^6}{2^2 \cdot 4^2 \cdot 6^2} + \cdots \right) \, dx}

The integral I I above has a closed form. Find the value of this closed form.

Submit your answer as I 2 I^2 to 2 decimal places.


The answer is 2.72.

Kartik Sharma by Kartik Sharma , 21, India

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1 solution

Let I = ∫ 0 ∞ f ( x ) g ( x ) d x \displaystyle I=\int_{0}^{\infty} f(x)g(x)dx

f ( x ) = ∑ n = 1 ∞ ( − 1 ) n − 1 x 2 n − 1 ∏ k = 1 n − 1 2 k = x ∑ n = 1 ∞ ( − 1 ) n − 1 x 2 ( n − 1 ) 2 n − 1 ( n − 1 ) ! = x e − x 2 2 \displaystyle \color{#D61F06}{f(x)} = \color{#3D99F6}{\sum_{n=1}^{\infty} (-1)^{n-1}\frac{x^{2n-1}}{\prod_{k=1}^{n-1}2k} = x\sum_{n=1}^{\infty}(-1)^{n-1} \frac{x^{2(n-1)}}{2^{n-1}(n-1)!} = xe^{-\frac{x^2}{2}}}

g ( x ) = ∑ n = 0 ∞ x 2 n 4 n ( n ! ) 2 = I 0 ( x ) \displaystyle \color{#D61F06}{g(x)} = \color{#3D99F6}{\sum_{n=0}^{\infty} \frac{x^{2n}}{4^n(n!)^2} = I_0(x)}

Here I n ( x ) I_n(x) is the Modified Bessel Function of the First kind , but we'll evaluate the integral step by step without using a standard result of Bessel Functions.

I = ∫ 0 ∞ f ( x ) g ( x ) d x = ∑ n = 0 ∞ 1 4 n ( n ! ) 2 ∫ 0 ∞ x e − x 2 2 x 2 n d x \displaystyle I = \int_{0}^{\infty} f(x)g(x) dx = \sum_{n=0}^{\infty}\frac{1}{4^n(n!)^2} \int_{0}^{\infty}xe^{-\frac{x^2}{2}}x^{2n}dx

So, I = ∑ n = 0 ∞ 1 2 n ( n ! ) 2 ∫ 0 ∞ e − t t n d t ⏟ G a m m a F u n c t i o n = ∑ n = 0 ∞ Γ ( n + 1 ) 2 n Γ 2 ( n + 1 ) = ∑ n = 0 ∞ = ( 1 2 ) n n ! = e \displaystyle I = \sum_{n=0}^{\infty}\frac{1}{2^n(n!)^2} \underbrace{\color{#D61F06}{\int_{0}^{\infty} e^{-t}t^{n} dt}}_{Gamma Function} = \sum_{n=0}^{\infty} \frac{\Gamma(n+1)}{2^n\Gamma^2(n+1)} = \sum_{n=0}^{\infty}=\frac{(\frac{1}{2})^n}{n!} = \sqrt{e}

So, I 2 = e \displaystyle I^2 =\boxed{e}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

how 4 n 4^n become 2 n 2^n in the denominator ???

Kushal Bose - 4 years, 7 months ago

Use the substitution x 2 2 = t \displaystyle \frac{x^2}{2}=t

Aditya Narayan Sharma - 4 years, 7 months ago

This was Prob A3 on the 1997 William J. Putnam examination.....nice!

tom engelsman - 4 years, 6 months ago

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I see! How was it solved there?

Aditya Narayan Sharma - 4 years, 6 months ago

Nice solution. The only difference between your solution and mine is in the evaluation of f(x). I simply complicated it and used the general form of (2n)!!. But alas, landed on the same result.

Aditya Kumar - 4 years, 6 months ago

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