Evaluating the indefinite integral $\int \frac {x^7-1}{\log x}dx$ [duplicate]

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The integral to be evaluated is $$\int \frac {x^7-1}{\log x}dx$$

The actual question was to find the definite integral of the above in the limits $0$ to $1$. I could not apply any of the usual tricks to find that integral, hence I thought finding the indefinite integral first and then applying the limits would be the best way to evaluate the same.

I tried applying trigonometric substitution, which did'nt work. I also tried putting $x=e^t$, hoping to simplify the integral but that did not give me any results either.

How should I proceed to evaluate this integral?

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• just for reference, it evaluates to $\mathrm{Ei}(8 \log x) - \mathrm{li}(x) + c$ (shown here). – Dando18 May 19 '17 at 17:32
• What makes you believe it has a closed form? – user370967 May 19 '17 at 17:33
• The indefinite integral won't work. In the definite one, make the substitution $x=e^{-t}$ and look up "Frullani integral". The, you have derived what Mathematica said. – Professor Vector May 19 '17 at 17:35
• For reference, I believe the definite version of this is on one of MIT's integral competition files (solutions are also given) – user12345 May 19 '17 at 22:11
• @Bacon hence the "just for reference". Especially with integration, seeing the answer can help with outlining a solution. It's not meant to be a solution itself. – Dando18 May 22 '17 at 10:53

Since you asked about the indefinite integral, I will introduce a couple of identities and definitions:

$${\rm li} (x) = \int_0^x \frac{dt}{\ln t} \tag{1}$$

$$\operatorname{li}(x)=\operatorname{Ei}(\ln{x}) \tag{2}$$

$$\operatorname{Ei}(x)=-\int_{-x}^{\infty}\frac{e^{-t}}t\,dt. \tag{3}$$

Where ${\rm li} (x)$ is the logarithmic integral, and $\operatorname{Ei}(x)$ is the exponential integral.

Start by using the substitution you attempted, it is a good one: $$x=e^t \iff dx=e^t ~dt$$ Hence giving: $$\int \frac{x^7-1}{\ln{x}}~dx=\int \frac{e^{7t}-1}{t}\cdot e^t~dt=\int \frac{e^{8t}-e^t}{t}~dt=\color{green}{\int \frac{e^{8t}}{t}~dt}-\color{blue}{\int \frac{e^t}{t}~dt}$$ Substituting $u=8t$ on the $\color{green}{\text{green}}$ integral and applying $(3)$, we have: $$\int \frac{e^{8t}}{t}~dt=\int \frac{e^u}{u}~du=\operatorname*{Ei}(u)+c_1=\operatorname*{Ei}(8\ln{x})+c_1$$ The $\color{blue}{\text{blue}}$ integral is easy, we can apply $(3)$, then apply $(2)$. $$\int \frac{e^t}{t}~dt=\operatorname*{Ei}(t)+c_2=\operatorname*{Ei}(\ln{x})+c_2=\operatorname*{li}(x)+c_2$$ Combining the two results, we obtain the desired result. $$\bbox[5px,border:2px solid #C0A000]{\int \frac{x^7-1}{\ln{x}}~dx=\operatorname*{Ei}(8\ln{x})-\operatorname*{li}(x)+C}$$

Edit: Just to make the answer more slick, you can separate the integrals immediately: $$\int \frac{x^7-1}{\ln{x}}~dx=\color{#0066ff}{\int \frac{x^7}{\ln{x}}~dx}-\color{#990000}{\int \frac{1}{\ln{x}}~dx}$$ For the $\color{#0066ff}{\text{light blue}}$ integral, we can substitute $x=e^{t/8}$ to obtain: $$\int \frac{x^7}{\ln{x}}~dx=\int \frac{e^t}{t}~dt=\operatorname*{Ei}(t)+c_1=\operatorname*{Ei}(8\ln{x})+c_1$$ And for the $\color{#990000}{\text{red}}$ integral, we can just immediately apply $(1)$: $$\int \frac{1}{\ln{x}}~dx=\operatorname*{li}(x)+c_2$$ Combining the two results gives the same answer as the previous approach.

You said you originally wanted to solve the definite integral from $x = 0$ to $1$, so I will demonstrate how to do it without resorting to finding the primitive. To solve this integral, we use the Feynman trick of differentiating under the integral. First, consider

\begin{align} I(\alpha) &= \int_{0}^{1} \frac{x^{\alpha} - 1}{\log (x)} dx \quad (1) \end{align}

Differentiating with respect to $\alpha$ gives

\begin{align} I'(\alpha) &= \int_{0}^{1} \frac{d}{d \alpha} \left( \frac{x^{\alpha} - 1}{\log (x)} \right) dx \\ &=\int_{0}^{1} \frac{\log (x) \cdot x^{\alpha}}{\log (x)} dx \\ &= \int_{0}^{1} x^{\alpha} dx \\ &= \frac{1}{\alpha + 1} \end{align}

Integrating $I'$

\begin{align} I(\alpha) &= \int \frac{1}{\alpha + 1} d \alpha \\ &= \ln(\alpha + 1) + C \quad (2) \end{align}

Note that in $(1)$ we have $I(0) = 0$. Applying this to $(2)$, we find $C = 0$, so

\begin{align} I(\alpha) &= \ln(\alpha + 1) \\ \end{align}

Evaluating at $\alpha = 7$ gives the result.

• Thank's for providing this beautiful alternate approach! – Ananth Kamath May 23 '17 at 5:59