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The integral

$$\int_0^2 x^5 (2-x)^4 \text dx$$

Can be integrated using beta function by substituting $x=2\sin^{2}\theta$ and using the definition of the beta function

$$\beta(x,y)=2\int_{0}^{\frac{\pi}{2}}\sin^{2x-1}\theta\cos^{2y-1}\theta\, \text d\theta$$

Now suppose that our bounds have changed to $0$ and $1$. Is there a substitution that will make this new integral solvable by means of the beta function?

$$\int_0^1x^5(2-x)^4\text dx$$

abiessu
  • 8,115
MrDi
  • 1,001

2 Answers2

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This integral does not seem to be easily expressible in terms of a normal Beta Function. However, after the substitution $x\mapsto2x$, using the Incomplete Beta Function, it is $$ \begin{align} \int_0^1x^5(2-x)^4\,\mathrm{d}x &=2^{10}\int_0^{1/2}x^5(1-x)^4\,\mathrm{d}x\\[3pt] &=2^{10}\mathrm{B}\!\left(\tfrac12;6,5\right) \end{align} $$

We can use the Binomial Theorem and the identity $\mathrm{B}(m,n-1)=\frac{m+n-1}{n-1}\mathrm{B}(m,n)$ to get $$ \begin{align} \int_0^1x^5(2-x)^4\,\mathrm{d}x &=\int_0^1x^5(1+(1-x))^4\,\mathrm{d}x\\ &=\sum_{k=0}^4\binom{4}{k}\int_0^1x^5(1-x)^k\,\mathrm{d}x\\[6pt] &=\mathrm{B}(6,1)+4\,\mathrm{B}(6,2)+6\,\mathrm{B}(6,3)+4\,\mathrm{B}(6,4)+\mathrm{B}(6,5)\\[12pt] &=\left(\tfrac71\cdot\tfrac82\cdot\tfrac93\cdot\tfrac{10}4+4\cdot\tfrac82\cdot\tfrac93\cdot\tfrac{10}4+6\cdot\tfrac93\cdot\tfrac{10}4+4\cdot\tfrac{10}4+1\right)\mathrm{B}(6,5)\\[12pt] &=386\,\mathrm{B}(6,5) \end{align} $$

robjohn
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  • Would the downvoter care to comment? This integral does not seem to be expressible as a normal Beta Function, so I have shown that it is expressible as an Incomplete Beta Function. – robjohn Dec 10 '16 at 11:29
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With the listed integral, substitution and definition in hand, we have $x = 2\sin^2\theta,\text dx = 4\sin\theta\cos\theta\text { d}\theta$ from which we get

$$\int_0^2 x^5 (2-x)^4 \text dx = 2^{11}\int_{0}^{\frac{\pi}{2}}\sin^{11}\theta\cos^{9}\theta\, \text d\theta = 2^{10}\beta(6,5)$$

If the integral has a bounds-shift, and we wish to know what happens if the upper bound only goes to $1$, then (due to lack of symmetry of the function across the interval) we need to calculate

$$\int_1^2 x^5(2-x)^4\text dx = \int_0^1(1+y)^5(1-y)^4\text dy\\ =\int_0^1(1+y)(1-y^2)^4\text dy = \int_0^1(1-y^2)^4\text dy+\int_0^1y(1-y^2)^4\text dy\\ =\int_0^1(1-y^2)^4\text dy + \frac 12\beta(1,5)$$

Now, with $I = \int_0^1(1-y^2)^4\text dy = \int_1^2x^4(2-x)^4\text dx$ left, we have:

$$2I = \int_0^2x^4(2-x)^4\text dx = 2^9\beta(5,5)$$

Adding everything together, we get

$$\int_0^1x^5(2-x)^4\text dx = 2^{10}\beta(6,5)-2^8\beta(5,5)-\frac 12\beta(1,5)$$

If we manipulate this further using the factorial form, we find that

$$2^{10}\beta(6,5)-2^8\beta(5,5)-\frac 12\beta(1,5)\\ =2^{10}{5!4!\over 10!}-2^8{4!4!\over 9!}-{4!\over 2\cdot 5!}\\ =2^9{5!4!\over 10!}-2*7*9{5!4!\over 10!}=2\cdot 193\beta(6,5)$$

So there must exist a transformation such that

$$\int_0^1x^5(2-x)^4\text dx = 386\int_0^1u^5(1-u)^4\text du$$

This does not mean that such a transformation will be easy or obvious. Since $193=256-63=128+64+1$ is prime, it has only the above summation relationship to the original.

abiessu
  • 8,115
  • The last equation does not hold: $\int_0^1x^5(2-x)^4{\rm d}x \not= \int_0^1u(1-u)^4{\rm d}u$. Neither does the equality below "Adding everything together, we get" (from checking it numerically) – Winther Apr 01 '16 at 19:44
  • @Winther: I know, I'm trying to work out the correct details... – abiessu Apr 01 '16 at 19:47
  • The mistake is in the first $ -\beta(1,5)$. That integral is $\frac{1}{10}$ not $\beta(1,5)$. This leads to $\int_0^1 = 2^{10}\beta(6,5) - 2^8\beta(5,5) - \frac{1}{10}$. – Winther Apr 01 '16 at 19:50
  • @Winther: I have been suffering under the illusion that my mental calculations are "good enough" while simultaneously not double-checking them against real results. I am halting this practice. Thank you for taking the time to find the error in my calculations. – abiessu Apr 02 '16 at 19:26