What are some methods to evaluate $ \int_{0}^{1} \frac{t^3 - 1}{\ln(t)} \, dt$? How would one integrate:
$$\int_{0}^{1} \frac{t^3 - 1}{\ln(t)} \, dt \, ?$$
I know how to do it using $u$-substitution and functions like $\text{Ei}(x)$ and $\text{Li}(x)$, but we didn’t learn that in class yet.
 A: Its a classic example for a trick, also known as the Feynman integration method/ differentiation under the integral sign.
What you should do, its set up some function, $g(x)$ for example, as:
$$g(x) = \int_0 ^{1} \frac{t^x - 1}{\ln(t)} dt$$
Now, we should try and derive it, like so:
$$g'(x) = \int_0 ^{1} \frac{\partial}{\partial x}\frac{t^x - 1}{\ln(t)} dt$$
$$= \int_0 ^{1} \frac{\partial}{\partial x} \frac{t^x}{\ln(t)} - \frac{\partial}{\partial x} \frac{1}{\ln(t)} dt$$
$$=\int_0 ^{1} \frac{t^x \ln(t)}{\ln(t)} dt = \int_0 ^1 t^x dt$$
$$=\frac{t^{x+1}}{x+1} \bigg|_0 ^{1}\bigg. = \frac{1}{x+1}$$
Now, we need to compute the anti derivative of $g'(x)$, which is rather simple, it is $\ln \lvert x+1\rvert$.
Now, $g(3)$ is the original integral you tried to compute, thus, the answer is $\ln(4)$, you can check it using integral calculator.
A: A method that powers through the problem is the following. Note that
$$ \text{Ei}(z) = \ln z + \gamma + z + \frac{z^2}{4} + \frac{z^3}{18} + \mathcal{O}(z^4) $$
is the expansion about $z \to 0$. Also the Exponential integral has the specific values $\text{Ei}(0) = - \infty$ and $\text{Ei}(- \infty) = 0$. Now,
\begin{align}
I &= \int_{0}^{1} \frac{t^x -1}{\ln(t)} \, dt  \\
&= \int_{- \infty}^{0} \frac{e^{(x+1) \, u} - e^{u}}{u} \, du \hspace{10mm} \text{where} \:  t = e^{u} \\
&= \left[ \text{Ei}((x+1) \, u) - \text{Ei}(u) \right]_{- \infty}^{0} \\
&= \lim_{u \to 0} \, \left( \text{Ei}((x+1) \, u) - \text{Ei}(u) \right) \\
&= \lim_{u \to 0} \left( \ln(x+1) + x \, u + \mathcal{O}(u^2) \right) \\
&= \ln(x+1).
\end{align}
A second method is to expand $t^x$ into a power series.
\begin{align}
I &= \int_{0}^{1} \frac{t^x -1}{\ln(t)} \, dt \\
&= \int_{0}^{1} \left( e^{x \, \ln(t)} - 1\right) \frac{dt}{\ln(t)} \\
&= \sum_{n=1}^{\infty} \frac{x^n}{n!} \, \int_{0}^{1} \left( \ln(t) \right)^{n-1} \, dt \\
&= \sum_{n=1}^{\infty} \frac{x^n}{n!} \, \int_{0}^{\infty} e^{- u} \, (-u)^{n-1} \, du  \hspace{10mm} \text{where} \hspace{5mm} t = e^{-u} \\
&= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^n}{n!} \, \int_{0}^{\infty} e^{-u} \, u^{n-1} \, du \\
&= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^n}{n} \\
&= \ln(1+x).
\end{align}
The general result is then
$$ \int_{0}^{1} \frac{t^x -1}{\ln(t)} \, dt = \ln(x+1). $$
When $x = 3$ the integral becomes
$$ \int_{0}^{1} \frac{t^3 -1}{\ln(t)} \, dt = 2 \, \ln 2. $$
A: Using
$$ \lim_{n\to\infty}n(x^{\frac1n}-1)=\ln x $$
one has
\begin{eqnarray}
\int_0^1\frac{t^3-1}{\ln t}dt&=&\lim_{n\to\infty}\int_0^1\frac{t^3-1}{n(t^{\frac1n}-1)}dt\\
&=&\lim_{n\to\infty}\int_0^1\frac1n\frac{(t^3-1)\sum_{k=0}^{n-1}t^{\frac kn}}{t-1}dt\\
&=&\lim_{n\to\infty}\int_0^1\frac1n(t^2+t+1)\sum_{k=0}^{n-1}t^{\frac kn}dt\\
&=&\lim_{n\to\infty}\int_0^1\frac1n\sum_{k=0}^{n-1}\bigg(t^{\frac kn+2}+t^{\frac kn+1}+t^{\frac kn}\bigg)dt\\
&=&\lim_{n\to\infty}\frac1n\sum_{k=0}^{n-1}\bigg(\frac1{\frac kn+3}+\frac1{\frac kn+2}+\frac1{\frac kn+1}\bigg)dt\\
&=&\int_0^1\bigg(\frac1{x+3}+\frac1{x+2}+\frac1{x+1}\bigg)dx\\
&=&\ln4.
\end{eqnarray}
