# A challenging Integral

1. May 22, 2006

### benorin

Put

$$I_n(x)=\int \frac{n!dx}{x(x+1)\cdots (x+n)}$$​

The desired result is an evaluation of the sum

$$S_n(x)=\sum_{k=0}^{n}(-1)^{k}\left(\begin{array}{c}n\\k\end{array}\right) (x+k)\log (x+k)$$​

and so it is important that the evaluation of $$I_n(x)$$ be not by means of partial fraction decomposition, for this is already of note that $$S_n(x)$$ may be evaluated, that is to say: it is known to me that

$$\int\frac{n!dx}{x(x+1)\cdots (x+n)}=\int\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array}\right) \frac{(-1)^k}{x+k}dx = \sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array}\right) (-1)^{k}\log (x+k)+C$$​

whence integrating once again will produce an evaluation for $$S_n(x)$$.

I was working on developing a recurrence relation for $$I_n(x)$$ that I may solve it via generating functions; here of note is that it is also known to me that

$$I_{n+1}(x)=I_n(x)-I_n(x+1)$$

yet this is not useful to that end. What is desired is a non-homogeneous recurrence. Any thoughts?

Last edited: May 22, 2006
2. May 22, 2006

### Hurkyl

Staff Emeritus
How does S relate to I?

Is there any reason why you can't use the partial fraction decomposition, and then rearrage it into the form you desire?

3. May 22, 2006

### benorin

I made a mistake in my tex, now the three dots show up.

4. May 22, 2006

### benorin

$$\int I_n(x)dx = S_n(x) +g(x)$$

where I suspect that g(x) is rather tivial.

EDIT: in fact, g(x)=0.

Last edited: May 22, 2006
5. May 22, 2006

### nrqed

You don't really mean the integral of I_n(x), right? You mean I_n(x) = S_n(x), correct?

EDIT: Actually, I am a bit confused by the notation.I is an integral over x so why is it a function of x? Fo you mean $I_n(x) = \int_0^x \, f(x')$ or something to that effect?

Last edited: May 22, 2006
6. May 22, 2006

### nrqed

I haven't thought this one through completely but have you heard of Feynman's trick for combining propagators in quantum field theory. It *might* prove useful. It is the identity

$${ 1 \over A_1 A_2 \ldots A_n} = \int_0^1 dy_1 \ldots dy_n \, \delta (\sum_{i=1}^n y_i -1) \, { (n-1)! \over (y_1 A_1 + y_2 A_2 + \ldots y_n A_n)^n }$$

I just have a *gut* feeling that this might do the trick but I haven't worked out anything. It's 1 Am here and I am about to go to bed.

Patrick

7. May 23, 2006

### benorin

Because it is an indefinite integral, it is still a function of x after the integration is carried out, as in:$$\int \cos xdx=\sin x +C$$

8. May 23, 2006

### benorin

And no, I do mean to say that $$\int I_n(x)dx = S_n(x) +C$$ since

$$I_n(x)=\int\frac{n!dx}{x(x+1)\cdots (x+n)}=\int\sum_{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right) \frac{(-1)^k}{x+k}dx = \sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array }\right) (-1)^{k}\log (x+k)+C_1$$

we have

$$\int I_n(x)dx=\int\int\frac{n!dxdx}{x(x+1)\cdots (x+n)}= \int\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array }\right) (-1)^{k}\log (x+k) dx +\int C_1 dx$$
$$= \sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array }\right) (-1)^{k}\int\log (x+k) dx+C_1x = \sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array }\right) (-1)^{k}\left[ (x+k)\log (x+k)-(x+k)\right] +C_1x+C_2$$
$$=\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array }\right) (-1)^{k}(x+k)\log (x+k)-\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array }\right) (-1)^{k}(x+k)+C_1x+C_2+h_n(x)=S_n(x)+C$$

where a formula from the calculus of finite differences has been used to determine that

$$h_n(x)=\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array }\right) (-1)^{k}(x+k)=\left\{\begin{array}{cc}x,& \mbox{ if } n=0\\-1, & \mbox{ if } n=1\\0, & \mbox{ if } n\geq 1 \end{array}\right.$$​

Last edited: May 24, 2006
9. May 23, 2006

### benorin

An interesting development

An interesting development: recall that

$$\log x = \int_{0}^{\infty}\left( e^{-t}-e^{-xt}\right)\frac{dt}{t}$$​

so that (dropping the "+C") we have

$$I_n(x+1)=\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array}\right) (-1)^{k}\log (x+k+1)=\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array}\right) (-1)^{k}\int_{0}^{\infty}\left( e^{-t}-e^{-(x+k+1)t}\right)\frac{dt}{t}$$
$$= \int_{0}^{\infty}e^{-t}\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array}\right) (-1)^{k}\left( 1-e^{-(x+k)t}\right)\frac{dt}{t}=\int_{0}^{\infty}\left[e^{-t}\delta_{n,0}- e^{-t(x+1)}\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array}\right) \left(-e^{-t}\right) ^{k}\right] \frac{dt}{t}$$
$$=\int_{0}^{\infty}\left[e^{-t}\delta_{n,0}- e^{-t(x+1)}\left(1-e^{-t}\right) ^{n}\right] \frac{dt}{t}$$​

where $$\delta_{n,0} = \left\{\begin{array}{cc}1,&\mbox{ if } n=0\\0, & \mbox{ if } n\neq 0\end{array}\right.$$ is the discete delta function, hence the evaluation

$$I_n(x)=\left\{\begin{array}{cc}\int_{0}^{\infty}\left(e^{-t}- e^{-xt}\right) \frac{dt}{t}=\log x ,&\mbox{ if } n=0\\ -\int_{0}^{\infty}e^{-tx}\left(1-e^{-t}\right) ^{n}\frac{dt}{t}, & \mbox{ if } n\neq 0\end{array}\right.$$​

Last edited: May 24, 2006