Solve Tricky Definite Integral with x^a-1 Over ln(x) on Interval 0 to 1

[iAlexN]

I am trying to solve this integral:

$\int$ $\frac{x^a-1}{ln(x)} dx$ (with the interval from 0 to 1).

I have tried substitution but I could not find a way to get it to work. Any ideas on how to solve this?

Thanks!

 

[Simon Bridge]

$$\int_0^1 \frac{x^a-1}{\ln(x)} \;dx$$

In what context did this integral come up?
What sort of range is "a"?

Did you try splitting it up? $$\int_0^1 \frac{x^a}{\ln(x)} dx - \int_0^1 \frac{1}{\ln(x)} \;dx$$

Did you try substituting: $x = e^u$

Note: $$\int \frac{dx}{\ln(x)}=\text{li}(x)$$
http://mathworld.wolfram.com/LogarithmicIntegral.html

 

[lurflurf]

try first

$$\int_0^a \! x^t\, \mathrm{d}t$$

 

[benorin]

Integration Under the Integral Sign

$$\int_0^1 \frac{x^a-x^b}{\ln(x)}dx=\ln\left|\frac{b+1}{a+1}\right|$$

For proof, see http://mathworld.wolfram.com/IntegrationUndertheIntegralSign.html

 

[Saitama]

One way to approach this is by defining
$$I(a)=\int_0^1 \frac{x^a-1}{\ln x}$$
Differentiate both the sides wrt $a$ to obtain
$$I'(a)=\int_0^1 \frac{x^a\ln x}{\ln x}=\int_0^1 x^a\,dx$$
$$I'(a)=\frac{1}{a+1}$$
$$\Rightarrow I(a)=\ln(a+1)+C$$
Notice that $I(0)=0$, hence $C=0$.

$$\Rightarrow I(a)=\ln(a+1)$$