(P)i am so over this integral

Calculus Level 4

π ÷ ∫ 0 ∞ e − 3 x sin ⁡ ( x ) x d x = ? \large \pi \div \int_0^\infty \frac{e^{-\sqrt3 x} \sin(x)}{x} \, dx = \ ?


The answer is 6.

Rui-Xian Siew by Rui-Xian Siew , 24, Hong Kong

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2 solutions

Tanishq Varshney
Nov 15, 2015

Apply Laplace transformation and that's it!!

L ( sin ⁡ t t ) = ∫ s ∞ 1 s 2 + 1 d s \large{L \left(\frac{\sin t}{t}\right)=\displaystyle \int _{s}^{\infty} \frac{1}{s^2+1} ds}

L ( sin ⁡ t t ) = arccot ( s ) \large{L \left(\frac{\sin t}{t} \right)=\text{arccot}(s)} ............ ( 1 (1

Also

L ( sin ⁡ t t ) = ∫ 0 ∞ e − s t sin ⁡ t t d t \large{L \left(\frac{\sin t}{t}\right)=\displaystyle \int _{0}^{\infty} e^{-st}\frac{\sin t}{t} dt} .........….. ( 2 (2

From ( 1 (1 and ( 2 (2

∫ 0 ∞ e − s t sin ⁡ t t d t = arccot ( s ) \large{\displaystyle \int _{0}^{\infty} e^{-st}\frac{\sin t}{t} dt=\text{arccot}(s)}

Here s = 3 s=\sqrt{3}

So we get π 6 \large{\boxed{\frac{\pi}{6}}}

Another method can be by using differentiation under integral assuming

I ( a ) = ∫ 0 ∞ e a x sin ⁡ x x d x \large{I(a)=\displaystyle \int^{\infty}_{0} e^{ax}\frac{\sin x}{x}dx}

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I feel like this method is, as they say, like nuking an anthill.

Jake Lai - 5 years, 6 months ago

Can u give a link for Laplace transformation? I just don't know.

Shyambhu Mukherjee - 5 years, 6 months ago

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Here

Tanishq Varshney - 5 years, 6 months ago

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Thanks u very much.

Shyambhu Mukherjee - 5 years, 6 months ago
Jake Lai
Nov 16, 2015

Using the Maclaurin expansion sin ⁡ x = ∑ k = 0 ∞ ( − 1 ) k x 2 k + 1 ( 2 k + 1 ) ! \displaystyle \sin x = \sum_{k=0}^\infty \frac{(-1)^kx^{2k+1}}{(2k+1)!} , we can reform the given integral like so:

∫ 0 ∞ e − 3 x sin ⁡ x x d x = ∫ 0 ∞ e − 3 x ∑ k = 0 ∞ ( − 1 ) k x 2 k ( 2 k + 1 ) ! d x = ∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) ! ∫ 0 ∞ e − 3 x x 2 k d x \int_0^\infty e^{-\sqrt 3x}\frac{\sin x}{x} \ dx = \int_0^\infty e^{-\sqrt 3x}\sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{(2k+1)!} \ dx = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} \int_0^\infty e^{-\sqrt 3x}x^{2k} \ dx

By the substitution t = 3 x t = \sqrt 3x , we have

∫ 0 ∞ e − 3 x x 2 k d x = ∫ 0 ∞ e − t ( t 3 ) 2 k d t 3 = ( 1 3 ) 2 k + 1 ∫ 0 ∞ e − t t 2 k d t \int_0^\infty e^{-\sqrt 3x}x^{2k} \ dx = \int_0^\infty e^{-t}\left( \frac{t}{\sqrt 3} \right)^{2k} \ \frac{dt}{\sqrt 3} = \left( \frac{1}{\sqrt 3} \right)^{2k+1}\int_0^\infty e^{-t}t^{2k} \ dt

Since it is known that ∫ 0 ∞ e − t t n d t = n ! \displaystyle \int_0^\infty e^{-t}t^n \ dt = n! for nonnegative integers n n , we thus have

∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) ! ∫ 0 ∞ e − 3 x x 2 k d x = ∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) ! ( 1 3 ) 2 k + 1 ∫ 0 ∞ e − t t 2 k d t = ∑ k = 0 ∞ ( − 1 ) k ( 1 / 3 ) 2 k + 1 ( 2 k ) ! ( 2 k + 1 ) ! = ∑ k = 0 ∞ ( − 1 ) k ( 1 / 3 ) 2 k + 1 2 k + 1 \begin{aligned} \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} \int_0^\infty e^{-\sqrt 3x}x^{2k} \ dx &= \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} \left( \frac{1}{\sqrt 3} \right)^{2k+1}\int_0^\infty e^{-t}t^{2k} \ dt \\ &= \sum_{k=0}^\infty \frac{(-1)^k(1/\sqrt 3)^{2k+1}(2k)!}{(2k+1)!} \\ &= \sum_{k=0}^\infty \frac{(-1)^k(1/\sqrt 3)^{2k+1}}{2k+1} \end{aligned}

We recognise this familiar form as the series expansion of the arctangent, namely that tan ⁡ − 1 x = ∑ k = 0 ∞ ( − 1 ) k x 2 k + 1 2 k + 1 \displaystyle \tan^{-1} x = \sum_{k=0}^\infty \frac{(-1)^kx^{2k+1}}{2k+1} . Hence, we obtain

∑ k = 0 ∞ ( − 1 ) k ( 1 / 3 ) 2 k + 1 2 k + 1 = tan ⁡ − 1 1 3 = π 6 \sum_{k=0}^\infty \frac{(-1)^k(1/\sqrt 3)^{2k+1}}{2k+1} = \tan^{-1} \frac{1}{\sqrt 3} = \frac{\pi}{6}

and so

π ÷ ∫ 0 ∞ e − 3 x sin ⁡ x x d x = π ÷ π 6 = 6 \pi \div \int_0^\infty e^{-\sqrt 3x}\frac{\sin x}{x} \ dx = \pi \div \frac{\pi}{6} = \boxed{6}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Your solution was beautiful but I was stuck in recognizing the series of arctan how did u find that?

Righved K - 5 years, 6 months ago

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We have that d x d y d y d x = 1 \dfrac{dx}{dy} \dfrac{dy}{dx} = 1 from the chain rule. If we use x = tan ⁡ y x = \tan y then sec ⁡ 2 y d y d x = 1 \sec^2 y \dfrac{dy}{dx} = 1 and so

d d x tan ⁡ − 1 x = 1 sec ⁡ 2 tan ⁡ − 1 x = 1 1 + tan ⁡ 2 tan ⁡ − 1 x = 1 1 + x 2 \frac{d}{dx} \tan^{-1} x = \frac{1}{\sec^2 \tan^{-1} x} = \frac{1}{1+\tan^2 \tan^{-1} x} = \frac{1}{1+x^2}

which stems from the fact that sec ⁡ 2 x = 1 + tan ⁡ 2 x \sec^2 x = 1+\tan^2 x .

Now, using that 1 1 + x 2 = ∑ n = 0 ∞ ( − 1 ) n x 2 n \displaystyle \frac{1}{1+x^2} = \sum_{n=0}^\infty (-1)^nx^{2n} (geometric series), we have

∫ d d x tan ⁡ − 1 x d x = ∫ ∑ n = 0 ∞ ( − 1 ) n x 2 n d x = ∑ n = 0 ∞ ( − 1 ) n ∫ x 2 n d x = ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 2 n + 1 \int \frac{d}{dx} \tan^{-1} x \ dx = \int \sum_{n=0}^\infty (-1)^nx^{2n} \ dx = \sum_{n=0}^\infty (-1)^n \int x^{2n} \ dx = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}

But, on the other hand, using the fundamental theorem of calculus we also have

∫ d d x tan ⁡ − 1 x d x = tan ⁡ − 1 x + C \int \frac{d}{dx} \tan^{-1} x \ dx = \tan^{-1} x + C

where C C is a constant of integration. Equation the two expressions obtained, we get ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 2 n + 1 = tan ⁡ − 1 x + C \displaystyle \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1} = \tan^{-1} x + C , which when evaluated at x = 0 x = 0 gives 0 = 0 + C 0 = 0 + C , implying that C = 0 C = 0 .

Hence, we finally have

tan ⁡ − 1 x = ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 2 n + 1 \tan^{-1} x = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}

However, this is true only for ∣ − x 2 ∣ = x 2 < 1 |-x^2| = x^2 < 1 . Fortunately, using the fact that tan ⁡ − 1 ( 1 x ) = π 2 − tan ⁡ − 1 x \tan^{-1}\left(\dfrac{1}{x}\right) = \dfrac{\pi}{2} - \tan^{-1} x , we can extend the series obtained to all x ∈ R x \in \mathbb{R} .

Jake Lai - 5 years, 6 months ago

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