# Solve this integral of degree $2043$

Find the value of $$\frac{\displaystyle\int_0^1(x+1)^{1010}\:\:dx}{\displaystyle\int_0^1(x^{2043}+1)^{1010}\:\:dx}$$

I evaluated the value of the numerator as $$\frac{2^{1011}-1}{1011}$$ But can't do the same for the denominator.

Any help is greatly appreciated.

• Interesting question. No idea. I wonder if there is some clever way to evaluate the expression by combining the integrals somehow / without evaluating nuerator and denominator separately? Also, in doing the integrals separately, maybe some recurrence relation upon integration by parts? Aug 22, 2022 at 13:06
• I’ve found two methods, but both seem to require a computer to calculate a large sum or many iterations of a formula. Is this a viable option or would you rather a method which evaluates the integral by hand? Aug 22, 2022 at 13:31
• @Quanto how????? Aug 22, 2022 at 13:45
• @Quanto Your estimate isn't correct - the denominator is much larger than $1$. Consider the term $\int_0^1 {1010\choose2}x^{4086}\,dx\approx 125$. Aug 22, 2022 at 14:02
• @Quanto Your estimate is off by a magnitude of $10^{298}$. Wolfram evaluates the integral as $1.06455 \times 10^{298}$ Aug 22, 2022 at 14:06

## 2 Answers

I’ll demonstrate 2 methods to evaluate the bottom integral. It’s likely possible to evaluate the two answers in closed form, but I haven’t found one yet, so feel free to add on to this answer.

Method 1 - Expanding brackets:

We can use the binomial expansion to write the expression as,

$$\int_0^1 (x^{2043}+ 1)^{1010}dx = \int_0^1 ( \sum_{r=0}^{1010} ({1010 \choose r} x^{2043r})) dx$$

Switching the integral and sum gives,

$$\sum_{r=0}^{1010} (\int_0^1 ({1010 \choose r} x^{2043r})dx) = \sum_{r=0}^{1010} ({1010 \choose r} [\frac{x^{2043r + 1}}{2043r + 1}]_0^1) = \sum_{r=0}^{1010} ({1010 \choose r} \frac{1}{2043r + 1})$$

I’m currently unsure whether this sum has a simple exact form. Maybe considering a Taylor series expansion will yield a similar series. If anyone can find something like this then feel free to add on to this answer.

Method 2 - Reduction Formula:

Let’s define $$I_n = \int_0^1 (x^{2043} + 1)^n dx$$. Then we wish to evaluate $$I_{1010}$$. By performing integration by parts with $$u = (x^{2043} + 1)^n$$ and $$\frac{dv}{dx} = 1$$, we get

$$I_n = [x(x^{2043}+1)^n]_0^1 - 2043n \int_0^1 (x^{2043}+1)^{n-1}(x^{2043})dx$$

Evaluating the brackets and re writing the integral slightly we get

$$I_n = 2^n - 2043n \int_0^1 (x^{2043}+1)^{n-1}(x^{2043}+1-1) dx = 2^n - 2043n \int_0^1 (x^{2043}+1)^n - (x^{2043} + 1)^{n-1} dx$$

And now we can get the reduction formula,

$$I_n = 2^n - 2043nI_n + 2043nI_{n-1}$$

Which rearranges to

$$I_n = \frac{2043nI_{n-1} + 2^n}{2043n + 1}$$

With the initial condition that $$I_0 = 1$$. We could use repeated iterations to calculate $$I_{1010}$$, but to do it by hand would require some sort of trick, because I can’t find a closed formula for $$I_n$$ using this recurrence relation. If anyone can then feel free to add on to this answer.

Edit: Plugging the recurrence relation into Wolfram Alpha actually gives a closed formula, https://www.wolframalpha.com/input?i=solve+f%28n%29+%3D+%282043n+f%28n-1%29+%2B+2%5En%29+%2F+%282043n+%2B+1%29%3B+f%280%29+%3D+1. It seems unlikely that there is a simple answer, since the integral evaluates to a fraction with far too many digits.

Using the Gaussian hypergeometric function $$\int \left(x^p+1\right)^n \,dx=x \, _2F_1\left(-n,\frac{1}{p};1+\frac{1}{p};-x^p\right)$$ $$\int_0^1 \left(x^p+1\right)^n \,dx=\, _2F_1\left(-n,\frac{1}{p};1+\frac{1}{p};-1\right)$$ $$I_{n,p}=\frac{\displaystyle\int_0^1(x+1)^{n}\:\:dx}{\displaystyle\int_0^1(x^{p}+1)^{n}\:\:dx}=\frac{2^{n+1}-1}{(n+1)\, _2F_1\left(-n,\frac{1}{p};1+\frac{1}{p};-1\right)}$$

What is interesting to notice is that $$I_{n,2n+k}\sim \frac{2 \,i^{\frac{1}{n}}\, \left(2^{n+1}-1\right) n}{(n+1) \,B_{-1}\left(\frac{1}{2 n},n+1\right)}+k+O(k^2)$$ which, for $$n=1010$$ and $$k=23$$ gives $$2039.00$$ while the result is $$2038.96$$.