Special Funcs: Evaluating $$\int \frac{1}{\sqrt{x}\ln(x)}$$

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SUMMARY

The integral $$\int \frac{1}{\sqrt{x}\ln(x)}$$ does not have an elementary antiderivative, necessitating the use of special functions for evaluation. By substituting $u = \sqrt{x}$, the integral transforms into $$\int \frac{2}{\ln(u^2)} du$$. Further analysis reveals that using the substitution $x = e^u$ leads to the Exponential Integral Function, specifically $$\text{Ei}\left(\frac{u}{2}\right)$$. For definite evaluation, the discussion suggests exploring the integral from 2 to 3.

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  • Understanding of integral calculus and special functions.
  • Familiarity with the Exponential Integral Function, denoted as Ei.
  • Knowledge of substitution techniques in integration.
  • Basic concepts of logarithmic functions and their properties.
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  • Explore numerical methods for evaluating definite integrals, particularly in cases without elementary solutions.
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Amad27
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Hi,

Recently, I had stumbled across:

$$\int \frac{1}{\sqrt{x}\ln(x)}$$
Let $f(x) = \frac{1}{\sqrt{x}\ln(x)}$

I noticed there is no elementary antiderivative. I want to evaluate this using special functions, but as of right now, I would like some advice as I have no clue about special functions.

Let $u = \sqrt{x}$

$$ = \int \frac{2}{ln(u^2)} du$$

This is where it gets interesting, there is no elementary antiderivative, then how can we evaluate this in terms of special functions.

And AFTER THAT I want to experiment with

$$\int_{2}^{3} f(x) \,dx$$

thanks!
 
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Olok said:
Hi,

Recently, I had stumbled across:

$$\int \frac{1}{\sqrt{x}\ln(x)}$$
Let $f(x) = \frac{1}{\sqrt{x}\ln(x)}$

I noticed there is no elementary antiderivative. I want to evaluate this using special functions, but as of right now, I would like some advice as I have no clue about special functions.

Let $u = \sqrt{x}$

$$ = \int \frac{2}{ln(u^2)} du$$

This is where it gets interesting, there is no elementary antiderivative, then how can we evaluate this in terms of special functions.

And AFTER THAT I want to experiment with

$$\int_{2}^{3} f(x) \,dx$$

thanks!

Setting $\displaystyle x=e^{u}$ the integral becomes...

$\displaystyle \int \frac{d x}{\sqrt{x}\ \ln x} = \int \frac{e^{\frac{u}{2}}}{u}\ d u = \text{Ei}\ (\frac{u}{2}) + c\ (1)$

... so that your 'special function' is the Exponential Integral Function...

Kind regards

$\chi$ $\sigma$
 
Olok said:
Hi,

Recently, I had stumbled across:

$$\int \frac{1}{\sqrt{x}\ln(x)}$$
Let $f(x) = \frac{1}{\sqrt{x}\ln(x)}$

I noticed there is no elementary antiderivative. I want to evaluate this using special functions, but as of right now, I would like some advice as I have no clue about special functions.

Let $u = \sqrt{x}$

$$ = \int \frac{2}{ln(u^2)} du$$

This is where it gets interesting, there is no elementary antiderivative, then how can we evaluate this in terms of special functions.

And AFTER THAT I want to experiment with

$$\int_{2}^{3} f(x) \,dx$$

thanks!

Hello again Olok,

I suggest letting $u = \ln(x)$ instead, so that $x = e^u$ and $dx = e^u\, du$. Thus

$$ \int \frac{dx}{\sqrt{x}\ln(x)} = \int \frac{e^u\, du}{e^{u/2}\,u} = \int \frac{e^{u/2}}{u}\,du.$$

Letting $v = -u/2$, $dv/v = du/u$, and so

$$ \int \frac{e^{u/2}}{u}\, du = \int \frac{e^{-v}}{v}\, dv.$$

Although we're dealing with indefinite integrals, the last integral is closely related to the exponential integral,

$$\text{Ei}(x) = \int_x^\infty \frac{e^{-t}}{t}\, dt.$$
 
Hi,

So can we find the definite integral of that from

$2$ to $3$?
 

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