Analytical expression for $ \int_0^b \frac{(a+x)^k - a^k}{x}dx $ Can an analytical expression be found for the following definite integral:
$$ \int_0^b \frac{(a+x)^k - a^k}{x}dx $$
I have tried in Wolfram Alpha. Unfortunately, it ran out of free calculation time...
In my particular application, I am faced with the following constraints on the various terms:
$$ a,b \in [0,1] \,\, \text{ and } \,\, a = 1-b \,\, \text{ and } \,\, k \in \mathbb{Z}^+$$
It is possible to get the solution using a discrete sum (using the binomial theorem) or using a recursion. However, we are looking for closed-form expressions without sums or integrals. Not sure it is possible...any help would be much appreciated!
 A: $$\begin{align}f(x)
=&(a+x)^k-a^k\\
=&\sum_{n=0}^k{k\choose n}a^{k-n}x^n-a^k\\
=&\sum_{n=1}^k{k\choose n}a^{k-n}x^n
\end{align}$$

$$\frac{f(x)}{x}=\sum_{n=1}^k{k\choose n}a^{k-n}x^{n-1}$$

$$\begin{align}\int_0^b\frac{f(x)}{x}\,\mathrm dx
=&\int_0^b\sum_{n=1}^k{k\choose n}a^{k-n}x^{n-1}\,\mathrm dx\\
=&\sum_{n=1}^k{k\choose n}a^{k-n}\int_0^bx^{n-1}\,\mathrm dx\\
=&\sum_{n=1}^k{k\choose n}a^{k-n}\frac{b^n}{n}
\end{align}$$
A: PARTIAL ANSWER AND OBSERVATION
This is not a complete solution; indeed currently I see no way of arriving at a solution that doesn't involve sums, but it's too long for the comments so I'm leaving it as an answer.
However, I have noticed something that may be of use to someone else attempting to answer this questions.
From using the binomial theorem, we arrive at the fact that the integral is equivalent to
$$\sum_{n=1}^k\binom{k}{n}\frac{1}{n}(1-a)^na^{k-n}.$$
It occurred to me that this looks very much like the sum
$$\sum_{n=1}^k\binom{k}{n}(1-a)^na^{k-n}$$ which does have a closed form; it is equal to $1-a^k$.
If anyone can see a way of using this result I will be very interested, and very pleased :)
A: $\begin{array}\\
\int_0^b \frac{(a+x)^k - a^k}{x}dx
&=\int_0^b (\sum_{i=0}^{k-1} a^i(a+x)^{k-i-1})dx\\
&=\sum_{i=0}^{k-1} a^i\int_0^b (a+x)^{k-i-1}dx\\
&=\sum_{i=0}^{k-1} a^i\dfrac{(a+x)^{k-i}}{k-i}|_0^b\\
&=\sum_{i=0}^{k-1} a^i\dfrac{(a+b)^{k-i}-a^{k-i}}{k-i}\\
&=\sum_{i=0}^{k-1} a^i\dfrac{1-a^{k-i}}{k-i}
\qquad a+b=1\\
&=\sum_{i=0}^{k-1} \dfrac{a^i-a^k}{k-i}\\
&=\sum_{i=0}^{k-1} \dfrac{a^i}{k-i}-\sum_{i=0}^{k-1} \dfrac{a^k}{k-i}\\
&=\sum_{i=1}^{k} \dfrac{a^{k-i}}{i}-a^k\sum_{i=1}^{k} \dfrac1{i}\\
&=a^k\left(\sum_{i=1}^{k} \dfrac{a^{-i}}{i}-\sum_{i=1}^{k} \dfrac1{i}\right)\\
&=a^k\sum_{i=1}^{k} \dfrac{a^{-i}-1}{i}\\
\end{array}
$
A: I am going to write another answer as the other would be very long. You can use the Incomplete Beta function to define it too. Let's split it into two integrals:
$$I_1=\int_0^b\frac{(a+x)^k}{x}\,dx$$
with $u=ax$ we get $du=dx/a$ so:
$$\frac1a\int_0^{ab}\frac{a^k(1+u)^k}{u/a}\,du=a^k\int_0^{ab}\frac{(1+u)^k}{u}\,du$$
now let $u=v-1$ so $du=dv$ and we get:
$$a^k\int_1^{{1+ab}}\frac{v^k}{(v-1)}\,dv=-a^k\int_1^{1+ab}v^k(1-v)^{-1}\,dv$$
Now remember that the incomplete Beta function is defined as:
$$B(x;\alpha,\beta)=\int_0^xt^{\alpha-1}(1-t)^{\beta-1}\,dt$$

as for the second integral, you just have:
$$-a^k\int_0^b\frac1x\,dx$$
which should be fairly easy
A: Substituting $x=a(y-1)$ and using $a+b=1$ your integral is equal to
$$ I(a) = a^k \int_1^{\frac{1}{a}} \frac{y^k-1}{y-1}dy = a^k J_k(a^{-1})$$
where $$ J_k(z) = \int_1^{z} \frac{y^k-1}{y-1}dy $$
While we need it for $z>1$, it's easier to express it for $z<1$ and then define it for $z>1$ as the analytic continuation. For $z<1$ we have
\begin{align} J_k(z) &= -\int_z^1 \frac{1-y^k}{1-y}dy = \\ &= -\int_0^1 \frac{1-y^k}{1-y}dy + \int_0^z \frac{1}{1-y}dy - \int_0^z \frac{y^k}{1-y}dy = \\
&= -\int_0^1 \sum_{n=0}^{k-1} y^k dy - \ln(1-z) - \int_0^z y^k (1-y)^{-1} dy = \\
&= -\sum_{n=0}^{k-1} \frac{1}{k+1} - \ln(1-z) - B(z;k+1,0) = \\
&= -H_k - \ln(1-z) - B(z;k+1,0)\end{align}
where $H_k$ are the harmonic numbers and $B(z;k+1,0)$ is the incomplete beta function. While he beta function has a logarithmic singularity for $z\to 1$, it is cancelled by $-\ln(1-z)$ and the whole function can by extended to $z>1$.
