MHB Definite Integral challenge #4

[Saitama]

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

 

[I like Serena]

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

Can we use omega or alpha functions?

 

[Saitama]

Can we use omega or alpha functions?

Nope. :P

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

 

[Krizalid1]

Hint:

Spoiler

Integration by parts is a powerful tool.

 

[Saitama]

Hint:

Spoiler

Integration by parts is a powerful tool.

Hi Krizalid!

This isn't a homework problem, can you please complete your solution?

 

[Krizalid1]

I know, it's just that I know the solution but I want other peoply to try it. :D

 

[lfdahl]

Spoiler

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:

\[I = 2^{\beta-1}\frac{(\alpha!)^2(\alpha+1)(\alpha+1+\beta)(
\alpha+1+2\beta)...(\alpha+1+\alpha\beta)}{\beta^{\alpha}\alpha!(2\alpha+1)!} \\\\=2^{\beta-1}\frac{(\alpha+1)!\frac{3}{2}\frac{5}{2}\frac{7}{2}...\frac{2\alpha+1}{2}}{(2\alpha+1)!}=2^{\alpha+1}\frac{(\alpha+1)!(2\alpha+1)!}{(2\alpha+1)!2^{
\alpha}\alpha!}=2(\alpha+1)=\beta = 2010\]

 

[alyafey22]

ATTENTION :
lfdahl is using the prohibited beta :p.

 

[Saitama]

Here goes the solution without the use of weird functions. :p

(Suggested solution)

Spoiler

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,
$$2^{2009}\frac{N(1004)}{D(1004)}=2(1004+1)=\boxed{2010}$$

 

[I like Serena]

Nice solution!

(I miss the thanks button. ;))

 

[lfdahl]

Here goes the solution without the use of weird functions. :p

(Suggested solution)

Spoiler

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