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If $\displaystyle I = \int_{0}^{1}x^{1004}\cdot (1-x)^{1004}dx$ and $\displaystyle J=\int_{0}^{1}x^{1004}\cdot \left(1-x^{2010}\right)^{1004}dx\;,$

Then Relation between $I$ and $J.$

$\bf{My\; Try::}$ Given $$\displaystyle J = \int_{0}^{1}x^{1004}\cdot (1-x^{2010})^{1004}dx\;,$$ Now Put $\displaystyle x^{1005}=t\;,$ Then $1005x^{1004}dx = dt$

And Changing Limit, We get $$\displaystyle J=\frac{1}{1005}\int_{0}^{1}(1-t^2)^{1004}dt = \frac{1}{1005}\int_{0}^{1}\left[1-(1-t)^2\right]^{1004}dt$$

So we get $$\displaystyle J=\frac{1}{1005}\int_{0}^{1}t^{1004}\cdot (2-t)^{1004}dt$$

Now How can I solve after that , Help me

Thanks

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juantheron
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2 Answers2

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Using your substitution for $J$ we have

$$J=\frac{1}{1005}\int_{0}^{1}(1-t)^{1004}(1+t)^{1004}dt=\frac{1}{1005}\int_{0}^{1}t^{1004}(2-t)^{1004}dt$$

Now use substitution $t=2y$ we get

$$J=\frac{1}{1005}\int_{0}^{\frac{1}{2}}2^{2009}y^{1004}(1-y)^{1004}dy$$ Hence

$$J=\frac{2^{2009}}{1005}\int_{0}^{\frac{1}{2}}y^{1004}(1-y)^{1004}dy$$

But $$I=2\int_{0}^{\frac{1}{2}}y^{1004}(1-y)^{1004}dy$$

So $$\frac{I}{J}=\frac{2010}{2^{2009}}$$

Ekaveera Gouribhatla
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2

It may be more instructive to look at the general case. Define $$I(m,n) = \int_{x=0}^1 x^n (1-x)^m \, dx, \quad J(m) = \int_{x=0}^1 x^m (1-x^{2m+2})^m \, dx.$$ Then as you wrote with your choice of substitution $$u = x^{m+1}, \quad du = (m+1) x^m \, dx,$$ we find with the binomial theorem $$J(m) = \frac{1}{m+1} \int_{u=0}^1 (1-u^2)^m \, du = \frac{1}{m+1} \int_{u=0}^1 (1+u)^m (1-u)^m \, du = \frac{1}{m+1} \sum_{k=0}^m \binom{m}{k} I(m,k).$$ Now we observe through integration by parts with the choice $u = (1-x)^m$, $du = -m(1-x)^{m-1} \, dx$, $dv = x^n \, dx$, $v = \frac{1}{n+1} x^{n+1}$, that $$I(m,n) = \left[-\frac{1}{n+1} x^{n+1} (1-x)^m \right]_{x=0}^1 + \frac{m}{n+1} \int_{x=0}^1 x^{n+1} (1-x)^{m-1} \, dx = \frac{m}{n+1} I(m-1,n+1).$$ With the addition that $I(0,n) = \frac{1}{n+1}$ trivially, we can inductively show that $I(m,n)$ for nonnegative integers $m, n$ is given by $$I(m,n) = \frac{m! \, n!}{(m+n+1)!}.$$ Therefore, $$J(m) = \frac{1}{m+1} \sum_{k=0}^m \frac{m!}{k!(m-k)!} \frac{m! \, k!}{(m+k+1)!} = \frac{1}{m+1} \sum_{k=0}^m \frac{(m!)^2}{(m-k)! (m+k+1)!}$$ and $$\frac{J(m)}{I(m,m)} = \frac{1}{m+1} \sum_{k=0}^m \binom{2m+1}{m-k} = \frac{1}{m+1} \sum_{k=0}^m \binom{2m+1}{k} = \frac{4^m}{m+1},$$ the last identity of which I have left as a simple exercise.

heropup
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