Reduction formulae problem!!!!

Eh guys, recently i was practicing reduction formulae questions and i stumbled on a problem which adds more problem to me. :) I was unable to do it.

Find the reduction formulae of

InI_{n} = ∫0πe−xsin⁡nxdx\displaystyle\int_{0}^{\pi} e^{-x} \sin^{n}x dx

pls guys, help me!!!

#Calculus

Note by Samuel Ayinde
6 years, 2 months ago

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1 vote

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Comments

Here's the solution

Using Integration By Parts, we get

In=∫0πsin⁡nx e−x dxI_n = \int_0^\pi \sin^n x\ e^{-x}\ dx =(sin⁡nx∫e−x dx∣0π + n∫0πsin⁡n−1xcos⁡x e−x dx= \left(\sin^n x \int e^{-x}\ dx\right|_0^\pi \ + \ n\int_0^\pi \sin^{n-1}x \cos x\ e^{-x}\ dx =n∫0πsin⁡n−1xcos⁡x e−x dx= n\int_0^\pi \sin^{n-1}x \cos x\ e^{-x}\ dx

Again applying Integration By Parts

=n(sin⁡n−1xcos⁡x∫e−x dx∣0π + n∫0π((n−1)sin⁡n−2xcos⁡2x−sin⁡nx)e−x dx= n\left(\sin^{n-1} x \cos x \int e^{-x}\ dx\right|_0^\pi \ +\ n\int_0^\pi \left((n-1)\sin^{n-2} x \cos^2 x - \sin^n x\right)e^{-x}\ dx =n∫0π((n−1)sin⁡n−2x(1−sin⁡2x)−sin⁡nx)e−x dx= n\int_0^\pi \left((n-1)\sin^{n-2} x (1-\sin^2 x) - \sin^n x\right)e^{-x}\ dx =n∫0π((n−1)sin⁡n−2x−nsin⁡nx)e−x dx= n\int_0^\pi \left((n-1)\sin^{n-2} x - n\sin^n x\right)e^{-x}\ dx =n(n−1)∫0πsin⁡n−2x e−x dx−n2∫0πsin⁡nx e−x dx= n(n-1)\int_0^\pi \sin^{n-2} x\ e^{-x}\ dx - n^2 \int_0^\pi \sin^n x\ e^{-x}\ dx =n(n−1)In−2−n2In=In= n(n-1) I_{n-2} - n^2I_n = I_n

On simplifying, we get

In=n(n−1)n2+1In−2\boxed{I_n = \dfrac{n(n-1)}{n^2+1}I_{n-2}}

Now, we compute, I0I_0 and I1I_1 using successive IBP.

I0=∫0πe−x dx=1−e−πI_0 = \int_0^\pi e^{-x}\ dx = 1-e^{-\pi}

and I1=∫0πe−xsin⁡x dx=∫0πe−xcos⁡x dx=(−cos⁡xe−x∣0π+∫0πsin⁡x e−x dx=(1+e−π)+I1⇒I1=1+e−π2\text{and}\ I_1 = \int_0^\pi e^{-x} \sin x\ dx = \int_0^\pi e^{-x} \cos x\ dx \\ = \left(-\cos x e^{-x} \right|_0^\pi + \int_0^\pi \sin x\ e^{-x}\ dx = (1+e^{-\pi}) + I_1 \\ \Rightarrow I_1 = \dfrac{1+e^{-\pi}}{2}

So, can you guess, why did we compute I0I_0 and I1I_1?

It's because if nn is even then -

In=n(n−1)n2+1In−2=n(n−1)n2+1(n−2)(n−3)(n−2)2+1In−4=⋯=n(n−1)n2+1(n−2)(n−3)(n−2)2+1…I0I_n = \dfrac{n(n-1)}{n^2+1}I_{n-2} = \dfrac{n(n-1)}{n^2+1}\dfrac{(n-2)(n-3)}{(n-2)^2+1}I_{n-4} =\cdots = \dfrac{n(n-1)}{n^2+1}\dfrac{(n-2)(n-3)}{(n-2)^2+1}\ldots I_0

and similarly if nn is odd, then

In=n(n−1)n2+1(n−2)(n−3)(n−2)2+1…I1I_n = \dfrac{n(n-1)}{n^2+1}\dfrac{(n-2)(n-3)}{(n-2)^2+1}\ldots I_1

As an example, we try computing the integral for n=2,3,…n=2,3,\ldots

I2=25I0=25(1−e−π)I_2 = \frac{2}{5}I_0 = \frac{2}{5}(1-e^{-\pi})

I3=35I1=310(1+e−π)I_3 = \frac{3}{5}I_1 = \frac{3}{10}(1+e^{-\pi})

I4=817I2=1685(1−e−π)I_4 = \frac{8}{17}I_2 = \frac{16}{85}(1-e^{-\pi})

And now, similarly, you can compute the integral for any n∈Nn \in N

Note: I am not sure if an explicit solution to the above recurrence, that we found, exists or not and I am still working on it's solution. But W|A gives the solution to that recurrence formula as

In=2πc1csch(π)Γ(n)Γ(n+1)Γ(n+1−i)Γ(n+1+i)I_n = \dfrac{2\pi c_1 \text{csch}(\pi) \Gamma(n) \Gamma(n+1)}{\Gamma(n+1-i) \Gamma(n+1+i)}

Kishlaya Jaiswal - 6 years, 2 months ago

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thanks so much. initially i used integration by parts but i got confused at a point. Thanks once again.

samuel ayinde - 6 years, 2 months ago

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It was all mine pleasure.

Kishlaya Jaiswal - 6 years, 2 months ago
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