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# 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

$$I_{n}$$ = $$\displaystyle\int_{0}^{\pi} e^{-x} \sin^{n}x dx$$

pls guys, help me!!!

Note by Samuel Ayinde
2 years, 10 months ago

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Here's the solution

Using Integration By Parts, we get

$I_n = \int_0^\pi \sin^n x\ e^{-x}\ dx$ $= \left(\sin^n x \int e^{-x}\ dx\right|_0^\pi \ + \ n\int_0^\pi \sin^{n-1}x \cos x\ e^{-x}\ dx$ $= n\int_0^\pi \sin^{n-1}x \cos x\ e^{-x}\ dx$

Again applying Integration By Parts

$= n\left(\sin^{n-1} x \cos x \int e^{-x}\ dx\right|_0^\pi \ +\ n\int_0^\pi \left((n-1)\sin^{n-2} x \cos^2 x - \sin^n x\right)e^{-x}\ dx$ $= n\int_0^\pi \left((n-1)\sin^{n-2} x (1-\sin^2 x) - \sin^n x\right)e^{-x}\ dx$ $= n\int_0^\pi \left((n-1)\sin^{n-2} x - n\sin^n x\right)e^{-x}\ dx$ $= n(n-1)\int_0^\pi \sin^{n-2} x\ e^{-x}\ dx - n^2 \int_0^\pi \sin^n x\ e^{-x}\ dx$ $= n(n-1) I_{n-2} - n^2I_n = I_n$

On simplifying, we get

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

Now, we compute, $$I_0$$ and $$I_1$$ using successive IBP.

$I_0 = \int_0^\pi e^{-x}\ dx = 1-e^{-\pi}$

$\text{and}\ I_1 = \int_0^\pi e^{-x} \sin x\ dx = \int_0^\pi e^{-x} \cos x\ dx \\ = \left(-\cos x e^{-x} \right|_0^\pi + \int_0^\pi \sin x\ e^{-x}\ dx = (1+e^{-\pi}) + I_1 \\ \Rightarrow I_1 = \dfrac{1+e^{-\pi}}{2}$

So, can you guess, why did we compute $$I_0$$ and $$I_1$$?

It's because if $$n$$ is even then -

$I_n = \dfrac{n(n-1)}{n^2+1}I_{n-2} = \dfrac{n(n-1)}{n^2+1}\dfrac{(n-2)(n-3)}{(n-2)^2+1}I_{n-4} =\cdots = \dfrac{n(n-1)}{n^2+1}\dfrac{(n-2)(n-3)}{(n-2)^2+1}\ldots I_0$

and similarly if $$n$$ is odd, then

$I_n = \dfrac{n(n-1)}{n^2+1}\dfrac{(n-2)(n-3)}{(n-2)^2+1}\ldots I_1$

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

$I_2 = \frac{2}{5}I_0 = \frac{2}{5}(1-e^{-\pi})$

$I_3 = \frac{3}{5}I_1 = \frac{3}{10}(1+e^{-\pi})$

$I_4 = \frac{8}{17}I_2 = \frac{16}{85}(1-e^{-\pi})$

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

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

$I_n = \dfrac{2\pi c_1 \text{csch}(\pi) \Gamma(n) \Gamma(n+1)}{\Gamma(n+1-i) \Gamma(n+1+i)}$

- 2 years, 10 months ago

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

- 2 years, 10 months ago

It was all mine pleasure.

- 2 years, 10 months ago