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Pattern – 36 : Definite integration

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⋆ Pattern – 36

I_n=\int_{0}^{\infty}{x^n\ e^{-x}\ dx=n!}

Proof :

Here, I_n=\ \int_{0}^{\infty}x^n\ e^{-x}dx

=\ \left[x^n\int{e^{-x}\ dx}\right]^\infty-\int\left(\frac{d}{dx}\ x^n\int^{0}{e^{-x}\ dx}\right)dx

=\ \left[\frac{x^n\ e^{-x}}{-1}\right]_0^\infty-\int{n\times \ \frac{x^{n-1}\times e^{-x}}{-1}}dx

=\ \left(0-0\right)+n\ \int{x^{n-1}e^{-x}\ dx}

=nI_{n-1}---\left(1\right)

Now,

I_n=n\times\left(n-1\right)I_{n-2}

=n\times\left(n-1\right)\left(n-2\right)I_{n-3}

…

=n\left(n-1\right)\left(n-2\right)... I_0---\left(2\right)

Now,

I_0=\ \int_{0}^{\infty}{e^{-x}\ dx}

=\ \left[\frac{e^{-x}}{-1}\right]_0^\infty

= – [ 0 – 1 ]

= 1

∴  from 2,

I_n = n(n-1)(n-2)…. 1

= n!

⋆ An Important Result

For ∀ x ∈ I,

x ∈ (n, n + 0.5)  ⟹ {x} < { – x}

x = 0.5  ⟹ {x} = { – x}

x ∈ (n + 0.5, n + 1)  ⟹ {x} > { – x}

Proof :

Let, x = 7, 2 ∈ (7, 7.5)

x : 7, 2  ⇒ [x] = 7

⇒ x – [x]= 7, 2.7

⇒ {x} = 0.2

x = 7.2  ⇒ -x : -7.2

⇒ [-x] = -8

⇒ -x – [ – x ]= 7.2-(-8)

⇒ { – x} = 0.8

{x} < {-x}

n + 0.5 :

x = 0.5 ∈ (0, 1)

x = 0.5  ⇒ [x] = [0,5] = 0

⇒ x – [x] = 0.5 – 0

⇒ {x} = 0.5

-x = -0.5  ⇒ [-x] = -1

⇒ x – [-x] = -0.5 + 1

⇒ {-x} = 0.5

(n+ 0.5, n+1) :

x = 7.7 ∈ (7.5 , 8)

x = 7.7  ⇒ [x] = 7

⇒ x – [x] = 7.7-7

⇒ {x} = 0.7

-x = -7.7 ⇒ [-x] = -8

⇒ -x –[-x]=-7.7 + 8

⇒ {-x} = 0.3

{x} > {-x}

 

Thank you…

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