MHB Definite Integral challenge #4

AI Thread Summary
The discussion revolves around evaluating the expression involving definite integrals without using beta or gamma functions. Participants emphasize the challenge of solving the problem using only elementary methods, with hints provided to guide others. There is a light-hearted exchange about the use of other functions, like omega or alpha, which are also deemed inappropriate. One user expresses a desire for others to attempt the solution rather than providing it outright. Overall, the thread highlights a collaborative effort to tackle a mathematical challenge while adhering to specific constraints.
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Evaluate:

$$2^{2009}\frac{\displaystyle \int_0^1 x^{1004}(1-x)^{1004}\,dx}{\displaystyle \int_0^1x^{1004}(1-x^{2010})^{1004}\,dx}$$

...of course without the use of beta or gamma functions. :p
 
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Pranav said:
...of course without the use of beta or gamma functions. :p

Can we use omega or alpha functions? :rolleyes:
 
I like Serena said:
Can we use omega or alpha functions? :rolleyes:

Nope. :P

I should have said that the problem is to be solved by elementary approaches.
 
Hint:

Integration by parts is a powerful tool.
 
Krizalid said:
Hint:

Integration by parts is a powerful tool.

Hi Krizalid!

This isn't a homework problem, can you please complete your solution? :rolleyes:
 
I know, it's just that I know the solution but I want other peoply to try it. :D
 
Let $\alpha = 1004$ and $\beta=2010$
I will try to calculate the integral:
\[I = 2^{\beta -1}\frac{\int_{0}^{1}x^\alpha (1-x)^\alpha dx}{\int_{0}^{1}x^\alpha (1-x^\beta )^\alpha dx}\]
Calculating the nominator:
\[\int_{0}^{1}x^\alpha (1-x)^\alpha dx = \left [ \frac{1}{\alpha +1}x^{\alpha +1}(1-x)^\alpha \right ]_{0}^{1}+\frac{\alpha }{\alpha +1}\int_{0}^{1}x^{\alpha+1}(1-x)^{\alpha -1}dx \\\\=\frac{\alpha }{\alpha +1}\int_{0}^{1}x^{\alpha+1}(1-x)^{\alpha -1}dx=\frac{\alpha}{\alpha+1}\left ( \left [ \frac{1}{\alpha+2}x^{\alpha+2}(1-x)^{\alpha-1} \right ]_{0}^{1} +\frac{\alpha-1}{\alpha+2}\int_{0}^{1}x^{\alpha+2}(1-x)^{\alpha-2}dx\right) \\\\=\frac{\alpha(\alpha-1)}{(\alpha+1)(\alpha+2)}\int_{0}^{1}x^{\alpha+2}(1-x)^{\alpha-2}dx = ... = \frac{\alpha!}{(\alpha+1)(\alpha+2)...(2\alpha+1)}=\frac{(\alpha!)^2}{(2\alpha+1)!}\]

Calculating the denominator:
\[\int_{0}^{1}x^\alpha (1-x^\beta)^\alpha dx = \left [ \frac{1}{\alpha +1}x^{\alpha +1}(1-x^\beta)^\alpha \right ]_{0}^{1}+\frac{\beta\alpha }{\alpha +1}\int_{0}^{1}x^{\alpha+\beta}(1-x^\beta)^{\alpha -1}dx \\\\=\frac{\beta\alpha }{\alpha +1}\int_{0}^{1}x^{\alpha+\beta}(1-x^\beta)^{\alpha -1}dx=\frac{\beta\alpha}{\alpha+1}\left ( \left [ \frac{1}{\alpha+\beta+1}x^{\alpha+\beta+1}(1-x^\beta)^{\alpha-1} \right ]_{0}^{1} +\frac{\beta(\alpha-1)}{\alpha+\beta+1}\int_{0}^{1}x^{\alpha+2\beta}(1-x)^{\alpha-2}dx\right) \\\\=\frac{\beta^2\alpha(\alpha-1)}{(\alpha+1)(\alpha+1+\beta)}\int_{0}^{1}x^{
\alpha+2\beta}(1-x^\beta)^{\alpha-2}dx\\\\ = \frac{\beta^3\alpha(\alpha-1)(\alpha-2)}{(\alpha+1)(\alpha+1+\beta)(\alpha+1+2\beta)} \int_{0}^{1}x^{\alpha+3\beta}(1-x^\beta)^{\alpha-3}dx= ... = \frac{\beta^{\alpha}\alpha!}{(\alpha+1)(\alpha+1+ \beta)(\alpha+1+2\beta)...(\alpha+1+\alpha\beta)}\]

Thus, the integral can be written:

\[I = 2^{\beta-1}\frac{(\alpha!)^2(\alpha+1)(\alpha+1+\beta)( \alpha+1+2\beta)...(\alpha+1+\alpha\beta)}{\beta^{\alpha}\alpha!(2\alpha+1)!}\]

$I$ can be reduced considerably:

\[I = 2^{\beta-1}\frac{(\alpha!)^2(\alpha+1)(\alpha+1+\beta)(
\alpha+1+2\beta)...(\alpha+1+\alpha\beta)}{\beta^{\alpha}\alpha!(2\alpha+1)!} \\\\=2^{\beta-1}\frac{(\alpha+1)!\frac{3}{2}\frac{5}{2}\frac{7}{2}...\frac{2\alpha+1}{2}}{(2\alpha+1)!}=2^{\alpha+1}\frac{(\alpha+1)!(2\alpha+1)!}{(2\alpha+1)!2^{
\alpha}\alpha!}=2(\alpha+1)=\beta = 2010\]
 
Last edited:
ATTENTION :
lfdahl is using the prohibited beta :p.
 
Here goes the solution without the use of weird functions. :p

(Suggested solution)

Let
$$N(k)=\int_0^1 x^k(1-x)^k\,dx \,\,\,(*)$$
and
$$D(k)=\int_0^1x^k(1-x^{2k+2})^k\,dx\,\,\,(**)$$
Use the substitution $x^{k+1}=t$ to obtain:
$$D(k)=\frac{1}{k+1}\int_0^1 (1-t^2)^k\,dt$$
$$\Rightarrow \frac{1}{2(k+1)}\int_{-1}^1 (1-t^2)^k\,dt=\frac{2^{2k-1}}{k+1}\int_{-1}^1 \left(\frac{1-x}{2}\right)^k\left(\frac{1+x}{2}\right)^k\,dx$$
Next, use the substitution,
$$\frac{1+x}{2}=u \Rightarrow dx=2du$$
to obtain:
$$D(k)=\frac{2^{2k}}{k+1}\int_0^1 u^k(1-u)^k\,du=\frac{2^{2k}}{k+1} \int_0^1 x^k(1-x)^k\,dx$$
$$D(k)=\frac{2^{2k}}{k+1}N(k)$$
Hence,
$$2^{2k+1}\frac{N(k)}{D(k)}=2(k+1)$$
Since $k=1004$, hence,
$$2^{2009}\frac{N(1004)}{D(1004)}=2(1004+1)=\boxed{2010}$$
 
  • #10
Nice solution!

(I miss the thanks button. ;))
 
  • #11
Pranav said:
Here goes the solution without the use of weird functions. :p

(Suggested solution)

Let
$$N(k)=\int_0^1 x^k(1-x)^k\,dx \,\,\,(*)$$
and
$$D(k)=\int_0^1x^k(1-x^{2k+2})^k\,dx\,\,\,(**)$$
Use the substitution $x^{k+1}=t$ to obtain:
$$D(k)=\frac{1}{k+1}\int_0^1 (1-t^2)^k\,dt$$
$$\Rightarrow \frac{1}{2(k+1)}\int_{-1}^1 (1-t^2)^k\,dt=\frac{2^{2k-1}}{k+1}\int_{-1}^1 \left(\frac{1-x}{2}\right)^k\left(\frac{1+x}{2}\right)^k\,dx$$
Next, use the substitution,
$$\frac{1+x}{2}=u \Rightarrow dx=2du$$
to obtain:
$$D(k)=\frac{2^{2k}}{k+1}\int_0^1 u^k(1-u)^k\,du=\frac{2^{2k}}{k+1} \int_0^1 x^k(1-x)^k\,dx$$
$$D(k)=\frac{2^{2k}}{k+1}N(k)$$
Hence,
$$2^{2k+1}\frac{N(k)}{D(k)}=2(k+1)$$
Since $k=1004$, hence,
$$2^{2009}\frac{N(1004)}{D(1004)}=2(1004+1)=\boxed{2010}$$

Nice solution indeed! I also miss the thanks-button
 
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