[Daily puzzle] Integral \ stacker news ~science

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684 sats \ 11 comments \ @south_korea_ln 31 Oct 2024 science

Need to work on some other stuff soon, so falling back on this one I had prepared a while ago for such an occasion.

Can you solve this integral?

 $\int_0^1 \frac{4x^3 \left(1 + x^{4 (2006)}\right)}{\left(1 + x^4\right)^{2008}} \, dx =\ ?$

Previous iteration: #746521 (answer in #746671).

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101 sats \ 2 replies \ @south_korea_ln OP 2 Nov 2024

First, break the integral into two parts as

 $\int_0^1 \frac{4x^3}{(1 + x^4)^{2008}} \, dx + \int_0^1 \frac{4x^3 \cdot x^{4(2006)}}{(1 + x^4)^{2008}} \, dx.$

Using the

u = 1 + x^4

substitution, the first integral becomes

 $\int_1^2 \frac{1}{u^{2008}} \, du = \frac{1}{2007}(1 - 2^{-2007}),$

while the second becomes

 $\int_1^2 \frac{(u - 1)^{2006}}{u^{2008}} \, du = \int_1^2 \frac{(1 - 1/u)^{2006}}{u^2} \, du.$

With the additional substitution

v = 1 - 1/u

, the final integral above becomes

 $\int_0^{1/2} v^{2006} \, dv = \frac{1}{2007}(2^{-2007})$

so that the original integral is equal to

 $\frac{1}{2007}(1 - 2^{-2007}) + \frac{1}{2007}(2^{-2007}) = \frac{1}{2007}.$

0 sats \ 1 reply \ @SimpleStacker 2 Nov 2024

Curious as to how the hint was related

21 sats \ 0 replies \ @south_korea_ln OP 2 Nov 2024

Ah i was trying to hint at the fact one had to do another substitution. You had already done one, i was trying to tell you to do another one. Instead of switching to integration by parts. Maybe not the best hint ;)

300 sats \ 2 replies \ @SimpleStacker 31 Oct 2024

This is as far as I got. Maybe someone else can take it the rest of the way

 $u = x^4$

 $du = 4x^3 dx$

 $\int_{0}^{1} \frac{4x^3 \left(1+x^{4(2006)}\right)}{(1+x^4)^{2008}} dx = \int_{0}^{1} \frac{1+u^{2006}}{(1+u)^{2008}} du$

 $= \int_{0}^{1} \frac{1}{(1+u)^{2008}} du + \int_{0}^{1} \frac{u^{2006}}{(1+u)^{2008}} du$

The left term can be easily calculated.

For the right term, let

 $F(k) = \int_{0}^{1} \frac{u^{k}}{(1+u)^{k+2}} du$

Using integration by parts, you can show that:

 $F(k) = \left. -\frac{u^{k}}{(k+1)(1+u)^{k+1}} \right|_{0}^{1} + \frac{k}{k+1} F(k-1)$

Starting at

F(0)

which is easy to calculate, you can use the above algorithm to eventually calculate

F(2006)

to find the answer.

0 sats \ 1 reply \ @south_korea_ln OP 1 Nov 2024

For the right term, one can do another substitution. Will write it out later.

0 sats \ 0 replies \ @south_korea_ln OP 2 Nov 2024

See here: #750702

0 sats \ 0 replies \ @south_korea_ln OP 31 Oct 2024

Hint: never change a winning team...

0 sats \ 0 replies \ @0xbitcoiner 31 Oct 2024

I bet it's greater than zero and less than one! Ahhhah

0 sats \ 2 replies \ @Aardvark 31 Oct 2024

I'm sticking with 42. It's going to be correct eventually.

10 sats \ 1 reply \ @south_korea_ln OP 31 Oct 2024

Just for you, I'll make sure that it'll happen, one day... just need to stick around, long enough

0 sats \ 0 replies \ @Aardvark 31 Oct 2024

I'll do my best!