Indefinite integral of sin(x)/ln(x)?

[romsofia]

Anyone how to integrate this? Integration by parts doesn't get me anywhere, and there isn't a maclaurin series for ln(x).

Thanks!

(This isn't a homework question, I just was randomly doing integrals)

 

[Char. Limit]

I can't remember if this counts as a Maclaurin series, but this is a series that sums to log(x):

- \sum_{n=1}^{\infty} \frac{\left(1-x\right)^n}{n}

 

[Mute]

Anyone how to integrate this? Integration by parts doesn't get me anywhere, and there isn't a maclaurin series for ln(x).

Thanks!

(This isn't a homework question, I just was randomly doing integrals)

What sorts of answers are acceptable to you? There is no answer in terms of standard elementary functions. (There isn't even an answer in terms of special functions according to wolframalpha).

\int dx~\frac{1}{\ln(x)}
itself is already a special function.

 

[romsofia]

I can't remember if this counts as a Maclaurin series, but this is a series that sums to log(x):

- \sum_{n=1}^{\infty} \frac{\left(1-x\right)^n}{n}

I don't think that counts as a maclaurin series :x, but thanks anyways! What I meant, is that's not the maclaurin series for ln(x) :x

What sorts of answers are acceptable to you? There is no answer in terms of standard elementary functions. (There isn't even an answer in terms of special functions according to wolframalpha).

\int dx~\frac{1}{\ln(x)}
itself is already a special function.

One without an integral still in it would be an ideal answer :x. May I ask how \int dx~\frac{1}{\ln(x)} is a special function, like where does it come up o.o?

 

[TylerH]

I can't remember if this counts as a Maclaurin series, but this is a series that sums to log(x):

- \sum_{n=1}^{\infty} \frac{\left(1-x\right)^n}{n}

The Taylor polynomials for log(1+x) only provide accurate approximations in the range −1 < x ≤ 1. Note that, for x > 1, the Taylor polynomials of higher degree are worse approximations.

Disappointing, right? Apparently, Taylor can't save use in all cases after all...

 

[micromass]

Disappointing, right? Apparently, Taylor can't save use in all cases after all...

Well, it can You just need to work in another base point. If you need to find Taylor approximations around 3, then you'd just have to use log(3+x).

Or, if that doesn't work, you can always use properties of logarithms to give you the correct answers:
For example, log(8)=3log(2) and log(2) has a good approximation.
Or, log(105)=log(105^11.(1/11))=11log(105^1/11) and this logarithm has a good approximation.

Isn't math cool?

 

[romsofia]

As much as I love taylor series (and I do), I don't think there is one for ln(x) or even log(x) (base 10)! Any other methods people can think of?

 

[micromass]

As much as I love taylor series (and I do), I don't think there is one for ln(x) or even log(x) (base 10)! Any other methods people can think of?

There certainly are Taylor series for ln(x) and log(x) around any point x>0. The problem is that the radius of convergence won't be entire R⁺. But how greater your center x, how greater your radius of convergence!

 

[romsofia]

There certainly are Taylor series for ln(x) and log(x) around any point x>0. The problem is that the radius of convergence won't be entire R⁺. But how greater your center x, how greater your radius of convergence!

Should have clarified my statement, I meant for them centered around zero :P aka maclaurin series

 

[Char. Limit]

Should have clarified my statement, I meant for them centered around zero :P aka maclaurin series

That's because log(x) itself isn't defined at zero. To have a Maclaurin series, a function needs to be defined (and, I believe, differentiable infinitely) at zero.

 

[romsofia]

That's because log(x) itself isn't defined at zero. To have a Maclaurin series, a function needs to be defined (and, I believe, differentiable infinitely) at zero.

Exactly, which is why we can't substitute ln(x) for a power series in the integral :x

 

[Char. Limit]

I don't even know how much a power series in the denominator would help anyway.

 

[romsofia]

Well it would've opened more options in terms of integrating :x

 

[pwsnafu]

"[URL to Wolfram
[/URL]

f(z) = \sum_{k=0}^{\infty} a_k z^k
g(z) = \sum_{k=0}^{\infty}b_k z^k; b_0 \neq 0

define
p_{j,0}=1
p_{j,k} = \frac{1}{b_0 k} \sum_{m=1}^k (jm+m-k) b_m p_{jk-m}

\frac{f(z)}{g(z)} = \frac{1}{b_0} \sum_{k=0}^\infty \sum_{j=0}^k (j+1) a_{k-j}\sum_{r=0}^j \frac{(-1)^{r}}{r+1} {j \choose r} p_{r,j} z^k

 

[Char. Limit]

"[URL to Wolfram
[/URL]

f(z) = \sum_{k=0}^{\infty} a_k z^k
g(z) = \sum_{k=0}^{\infty}b_k z^k; b_0 \neq 0

define
p_{j,0}=1
p_{j,k} = \frac{1}{b_0 k} \sum_{m=1}^k (jm+m-k) b_m p_{jk-m}

\frac{f(z)}{g(z)} = \frac{1}{b_0} \sum_{k=0}^\infty \sum_{j=0}^k (j+1) a_{k-j}\sum_{r=0}^j \frac{(-1)^{r}}{r+1} {j \choose r} p_{r,j} z^k

All right, now for a question. Is your quadruple sum there integrable?

 

[romsofia]

\frac{f(z)}{g(z)} = \frac{1}{b_0} \sum_{k=0}^\infty \sum_{j=0}^k (j+1) a_{k-j}\sum_{r=0}^j \frac{(-1)^{r}}{r+1} {j \choose r} p_{r,j} z^k

Is the {j \choose r} suppose to be a fraction...? Also, I'm completely lost from how you defined {sin(x)=f(z)} or {g(z)=1/ln(x)}

 

[romsofia]

I'm still interested if anyone has any other methods on a way to integrate this!

 

[Mute]

\int_0^x dt \frac{\sin t}{\ln t} = \int dt \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \frac{t^{2n+1}}{\ln t} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \int_0^x dt \frac{t^{2n+1}}{\ln t}

According to Wolfram alpha:

\int dt~\frac{t^{n}}{\ln t} = \mbox{Ei}((n+1)\ln t) + \mbox{constant},
where

\mbox{Ei}(x) = \int_{-\infty}^x dt \frac{e^t}{t}
From this it can be seen that as t tends to zero Ei((n+1)ln(t)) tends to zero as well.

Hence,

\int_0^x dt \frac{\sin t}{\ln t} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\mbox{Ei}((2n+2)\ln x).

Note, however, this result assumes that swapping the sum and integral was valid. It may not be, so this may at best be an asymptotic series.

 

[TylerH]

I'm still interested if anyone has any other methods on a way to integrate this!

In general, doing random integration problems will leave you asking this question A LOT. Do you have a calculus book? They're your best bet for not wasting your time on impossible problems.

 

[romsofia]

\int_0^x dt \frac{\sin t}{\ln t} = \int dt \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \frac{t^{2n+1}}{\ln t} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \int_0^x dt \frac{t^{2n+1}}{\ln t}

According to Wolfram alpha:

\int dt~\frac{t^{n}}{\ln t} = \mbox{Ei}((n+1)\ln t) + \mbox{constant},
where

\mbox{Ei}(x) = \int_{-\infty}^x dt \frac{e^t}{t}
From this it can be seen that as t tends to zero Ei((n+1)ln(t)) tends to zero as well.

Hence,

\int_0^x dt \frac{\sin t}{\ln t} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\mbox{Ei}((2n+2)\ln x).

Note, however, this result assumes that swapping the sum and integral was valid. It may not be, so this may at best be an asymptotic series.

Will edit this post as I will be reading it after dinner!

In general, doing random integration problems will leave you asking this question A LOT. Do you have a calculus book? They're your best bet for not wasting your time on impossible problems.

I am done with calculus, and I have a lot of time on hands in school so I am trying to solve random problems for fun :D.

 

[TylerH]

I am done with calculus, and I have a lot of time on hands in school so I am trying to solve random problems for fun :D.

Then maybe you should get your "random problems" from a textbook.

 

[romsofia]

Thank you Mute! Looks valid to me.

 

[Mute]

Well, keep in mind it might not be. I didn't check whether or not exchanging the integral with the sum actually gives me a convergent series (and if it did, I suppose it could still be incorrect).

The exponential integral function has asymptotics \mbox{Ei}(x) \sim e^x/x + \mathcal O (1/x), so for large n

\mbox{Ei}((2n+2)\ln x) \sim e^{2n\ln x}/(2n \ln x) = x^{2n}/(2n \ln x)
So, a typical term in the series at large n is

\frac{x^{2n}}{(2n+1)!n}
which tends to zero as n gets large, so it looks like the series stands a chance of actually being convergent.

We can also check that the large x asymptotics are consistent. Starting with the series for large x,

I(x) \sim \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \frac{x^{2n+2}}{(2n+2) \ln x}= \frac{1}{\ln x}\sum_{n=0}^\infty \frac{(-1)^nx^{2n+2}}{(2n+2)!}

Letting k = n + 1,

I(x) \sim \frac{1}{\ln x} \sum_{k=1}^\infty \frac{(-1)^{k-1}x^{2k}}{(2k)!} = \frac{1-\cos(x)}{\ln x}
where we identified the series as being -(cos(x)-1).

So, we can check the large x asymptotics directly, too:

I(x) = \int_0^x dt~\frac{\sin t}{\ln t} = x\int_0^1 dy~\frac{\sin(xy)}{\ln(xy)}
where we changed variables t = xy. Writing ln(xy) = ln(x) + ln(y) and noting that y is not larger than one, for large x we may neglect ln(y) in the denominator, giving

I(x) \sim \frac{x}{\ln x}\int_0^1 dy~\sin(xy) = \frac{x}{\ln x}\frac{1-\cos(x)}{x} = \frac{1-\cos(x)}{\ln x}

So, the asymptotics work out, like we expected they should. Note that this doesn't prove the series is convergent, because even if the swapping of integral and sum wasn't legal it could still give a valid asymptotic series. This is just for fun.