Approximation to this integral

[anonymous No.4742106]

Does anyone have an approximation to this integral?

[Anonymous No.4742198]

e^x

[Anonymous No.4742213]

>>4742106
count the primes between a and b

or ask ramanujan

[Anonymous No.4742227]

taylor expand it.

[Anonymous No.4742238]

>>4742213
>count the primes between a and b

This

[Anonymous No.4742243]

<span class="math">/int /frac{1}{ln(x)} dx = /int /frac{1}{u}*e^(-u) du[/spoiler]

Am i missing something?

[Anonymous No.4742245]

>>4742238
except that it doesn't work.

[Anonymous No.4742254]

>>4742245
>herp

http://en.wikipedia.org/wiki/Prime_Number_Theorem#Prime-counting_function_in_terms_of_the_logarithmi
c_integral

[Anonymous No.4742251]

>>4742243
<span class="math">\int \frac{1}{ln(x)} dx=\int \frac{1}{u} * \frac{1}{e^u} du[/spoiler]

[Anonymous No.4742268]

Make the substitution 1-u= x and use the Maclaurin expansion of ln(1-u).

If the interval is small, it is a good approximation to use a 1st or 2nd order expansion. If more is needed, try expand to more terms and integrate by doing a partial fractional decomposition.

[Anonymous No.4742273]

>>4742254
>derp
what if a and b are between 10 and 100? it only works for very large x you idiot.

[Anonymous No.4742274]

>>4742268
but the ln is on the denominator?

[Anonymous No.4742276]

>>4742273
almost all numbers are very large

and as OP didn't specify a and b...

[Anonymous No.4742280]

>>4742274
So?
ln(1-u) is real analytic so long as u is never equal to 1. It is identically equal to its Taylor expansion.

If your interval is such that u is ever 1`, then this method will not work.

[Anonymous No.4742281]

>>4742280

Clearly u can never be 0, either.

[Anonymous No.4742293]

>>4742280
>>4742280
pretty sure that the taylor expansion is about as bad as counting primes

ramanujan did some better aproximations

[Anonymous No.4742299]

Maybe expand in a Laurent series or something and use the calculus of residues to handle the singularities.

[Anonymous No.4742306]

just wolfram alpha this shit. seriously.

[Anonymous No.4742310]

http://en.wikipedia.org/wiki/Logarithmic_integral_function

[Anonymous No.4742338]

Let u = ln(ln(x)) so that du = (1/xln(x))dx.

Integrand becomes x du. x is exp{exp{u}} which can be Taylor expanded in e^u pretty nicely.

Pray the bounds aren't so bad.

[Anonymous No.4742345]

>>4742338

We can take this further. Let, v=e^u. Then the bounds which were ln(ln(a)) become ln(a) and the integrand is (e^v)/v .

That integral is well known, and integrated as a Maclaurin expansion.