Explicit form for $I(n) = \int_{0}^{1} \frac {(1-x)^n}{1+x^2} dx$

Question

Is it possible to find an explicit form for the Integral $$I(n) = \int_{0}^{1} \frac {(1-x)^n}{1+x^2} dx$$

for $n \ge 0$?

I tried differentiating under the Integral sign but it doesnt seem to help.

If we try to replace either $(1-x)^n$ or $\frac 1{1+x^2}$ with an infinite series representation, it gets messier. We can get that $I(n)$ is equal to $\sum_{k=0}^{\infty} (-1)^{k}\frac{n!(2k)!}{(2k+n+1)!}$ but I want a more explicit representation.

If I plug the Integrals into Wolfram alpha, I can see that there is some kind of pattern involving $\pi$ and $\ln(2)$. That should be intuitiv, because the infinite sum above is similar to

$$\sum_{k=1}^{\infty} \frac {(-1)^k}{k} = -\ln(2)$$

and

$$\sum_{k=1}^{\infty} \frac {(-1)^k}{2k-1} = -\frac{\pi}{4}$$

So my question is, is it possible to find a ecplicit form for $I(n)$, and does it involve $\pi$ and $\ln(2)$?

Answer 1

I'll sketch the basic strategy. Define $J(n)=\int_0^1\frac{x(1-x)^ndx}{1+x^2}$ so$$\int_0^1\frac{x^2(1-x)^ndx}{1+x^2}=\int_0^1(1-x)^ndx-I(n)=\frac{1}{n+1}-I(n).$$Hence$$I(n+1)=I(n)-J(n),\,J(n+1)=I(n)+J(n)-\frac{1}{n+1}.$$It helps to work with vectors and matrices: define$$M:=\left(\begin{array}{cc} 1 & -1\\ 1 & 1 \end{array}\right),\,u_n:=\left(\begin{array}{c} I\left(n\right)\\ J\left(n\right) \end{array}\right),\,v_n:=\left(\begin{array}{c} 0\\ \frac{1}{n+1} \end{array}\right)$$so$$u_0=\left(\begin{array}{c} \pi/4\\ \ln\sqrt{2} \end{array}\right),\,u_{n+1}=Mu_n-v_n\implies u_n=M^nu_0-\sum_{i=0}^{n-1}M^{n-1-i}v_i.$$The powers of $M$ are easy enough: they satisfy $M^{n+4}=-4M^n$ because$$M^2=\left(\begin{array}{cc} 0 & -2\\ 2 & 0 \end{array}\right),\,M^3=\left(\begin{array}{cc} -2 & -2\\ 2 & -2 \end{array}\right),\,M^4=-4I_2.$$

Answer 2

Using geometric series and Euler's Beta function

$$ I(n) = \int_{0}^{1}(1-x)^n\sum_{k\geq 0}(-1)^k x^{2k}\,dx =\sum_{k\geq 0}(-1)^k \frac{\Gamma(n+1)\Gamma(2k+1)}{\Gamma(n+2k+2)}=\sum_{k\geq 0}(-1)^k\frac{n!(2k)!}{(2k+n+1)!}.$$ On the other hand, by partial fraction decomposition $$ \frac{n!(2k)!}{(2k+n+1)!}=\frac{n!}{(2k+1)_{n+1}}=\sum_{h=0}^{n}\frac{(-1)^h \binom{n}{h}}{2k+h+1}$$ so $$ I(n) = \sum_{k\geq 0}(-1)^k \sum_{h=0}^{n}\frac{(-1)^h \binom{n}{h}}{2k+h+1}=\sum_{h=0}^{n}(-1)^h\binom{n}{h}\sum_{k\geq 0}\frac{(-1)^k}{2k+h+1} $$ is a linear combination of the tails of the series $$ \sum_{k\geq 0}\frac{(-1)^k}{2k+1}=\frac{\pi}{4}\qquad \sum_{k\geq 0}\frac{(-1)^k}{2k+2} = \frac{\log 2}{2}$$ as suspected.