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We know about the Fresnel Integrals:

$$C(x)=\int \cos x^2 \, dx,\quad S(x)=\int \sin x^2 \, dx$$

which can also be written as:

$$\int e^{ix^2}dx=C(x)+i\,S(x)$$

To make a more interesting and tetration based integral with a rapidly oscillating part of $\mathrm{Re(x)\ge0}$ integral converges. To make a more general result possible, let one consider the goal integral:

$$\int e^{ix^ x}dx=\int \cos\ x^x dx+i\,\int \sin\ x^x dx$$

Here is the graph over the real line of the integrand: enter image description here

Then the exponential Taylor Series comes to mind. The series converges for all x just like the expansion for $e^x$. We can integrate by each term to switch the sum and integral. Note the logarithm may also affect convergence. It uses the Exponential Integral function and Regularized Incomplete Gamma function:

$$\int e^{ix^x}dx=\int \sum_{n=0}^\infty\frac{i^n x^{nx}}{n!}dx=\sum_{n=0}^\infty \frac{i^n}{n!}\sum_{k=0}^\infty \frac{n^k}{k!}\int x^k \ln^k(x)dx\mathop=^{y=\ln(x)} \sum_{n=0}^\infty \frac{i^n}{n!}\sum_{k=0}^\infty \frac{n^k}{k!}\int e^{ky+1} y^k dy =C- \sum_{n=0}^\infty\sum_{k=0}^\infty \frac{i^nn^k\ln^{k+1}(x) \text E_{-k}(-(k+1)\ln(x))}{n!k!} = C+\sum_{n=0}^\infty\sum_{k=1}^\infty \frac{i^n(-n)^{k-1} Q(k, -k \ln(x)) \ln^{k-1}(x) \ln^{-(k+1)}(x) }{n! k^k}\mathop=^{\small{\ln^{k-1}(x) \ln^{-(k+1)}(x)=1}} C-\sum_{n=0}^\infty\sum_{k=1}^\infty \frac{i^n(-1)^k n^{k-1} Q(k, -k \ln(x)) }{n! k^k} $$

If this is correct, then how can I find the following? Note that when evaluating the bounds, the $”-\ln(0)”$ in the second argument Regularized Incomplete Gamma function will grow faster than the index of the summation, so it can be ignored:

$$\int_0^\infty e^{i x^x} dx= \int_0^\infty \cos\ x^x dx +i\,\int_0^\infty \sin\ x^xdx= -\lim_{x\to-\infty}\sum_{n=0}^\infty\sum_{k=1}^\infty \frac{(-1)^{k+\frac n2} n^{k-1} Q(k, kx) }{n! k^k} $$

Related question:

Can $$\int_0^\infty x^{-x} dx$$ be written without a limit?

Related answer which has an experimental solution:

Improper integral inequality including the golden ratio and $$\int_0^\infty \sin\left(x^{-x}\right)$$

Is there a good way to integrate this function and find this value? The limit makes the expression more complicated and perhaps we can write the answer without it. Please correct me and give me feedback!

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  • $\begingroup$ Have you looked at the definition of the incomplete gamma function. Because it seems that the limit diverges. $\endgroup$ Sep 15 at 3:49
  • $\begingroup$ @RounakSarkar This is from the $\int_0^\infty x^{-x} dx =1.995…$ question. Try small numbers and you will get this result. That is why I asked the question as the upper summation bound needs to be a lot larger than x to get the result, if there are no typos. $\endgroup$ Sep 15 at 11:40
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I am just going to focus on the indefinite integral, not the definite one.

Let,

$\textstyle\displaystyle{I=\int e^{ix^x}dx}$

Substitute, $u=x^x$ Then, $x=e^{W(\ln(x))}$ And $\textstyle\displaystyle{dx=\frac{du}{u(W(\ln(u))+1)}}$

This video will help you derive $dx$

Then,

$\textstyle\displaystyle{I=\int\frac{e^{iu}}{u}\frac{1}{W(\ln(u))+1}du}$

Now let,

$\textstyle\displaystyle{t(u)=\frac{1}{W(\ln(u))+1}}$

$\implies t^{-1}(u)=e^{\left(\frac{1}{u}-1\right)e^{\frac{1}{u}-1}}$

Now using the Lagrange Inversion Theorem we have,

$\textstyle\displaystyle{t(u)=1+\sum_{n=1}^{\infty}t_n\frac{(u-1)^n}{n!}}$

Where, $\textstyle\displaystyle{t_n=\lim_{w\rightarrow 1}\frac{d^{n-1}}{dw^{n-1}}\left[\left(\frac{w-1}{e^{\left(\frac{1}{w}-1\right)e^{\frac{1}{w}-1}}-1}\right)^n\right]}$

Here I have taken $a=1$.

As pointed out by Steven Clark and with the help of Tyma Gaidash in the comments, we can actually simplify $t_n$ using the Stirling number of the first kind. We have-

$\textstyle\displaystyle{t_n=(-1)^n\sum_{k=1}^{n}\left|S_{n}^{(k)}\right|k^k}$

$\textstyle\displaystyle{\left\{t_n\right\}_{n=1}^{\infty}}$

$\textstyle\displaystyle{=\{-1,5,-41,468,-6854,122582,-2589978,63129392,-1743732192,53827681152,-1836453542472,68620052332752,\dots\}}$

There is an OEIS sequence for $|t_n|$. (If you exclude the first term in the oeis sequence)

Using the Ratio test we can get the radius of convergence in terms of a limit of $t_n$. We have-

$\textstyle\displaystyle{\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|}$

$\textstyle\displaystyle{=\lim_{n\rightarrow\infty}\left|\frac{\frac{t_{n+1}(u-1)^{n+1}}{(n+1)!}}{\frac{t_n(u-1)^n}{n!}}\right|}$

$\textstyle\displaystyle{=|u-1|\lim_{n\rightarrow\infty}\left|\frac{t_{n+1}}{(n+1)t_n}\right|\lt 1}$

$\implies\textstyle\displaystyle{|u-1|\lt\lim_{n\rightarrow\infty}\left|\frac{(n+1)t_n}{t_{n+1}}\right|=L}$

If we let, $\textstyle\displaystyle{L_n=\left|\frac{(n+1)t_n}{t_{n+1}}\right|}$

$\textstyle\displaystyle{=\frac{(n+1)\sum_{k=1}^{n}\left|S_{n}^{(k)}\right|k^k}{\sum_{k=1}^{n+1}\left|S_{n+1}^{(k)}\right|k^k}}$

$L_{10}$$\approx 0.322417$

$L_{20}$$\approx 0.315299$

$L_{30}$$\approx 0.312843$

$L_{40}$$\approx 0.311598$

$L_{50}$$\approx 0.310846$

$L_{60}$$\approx 0.310343$

$\vdots$

The sequence $\{L_{n}\}_{n=1}^{\infty}$ seems to be steadily decreasing. Looking at the above approximations, it seems that $L\approx 0.31$

So we have,

$\textstyle\displaystyle{\frac{e^{iu}}{u}t(u)=\frac{e^{iu}}{u}+\sum_{n=1}^{\infty}\frac{t_n}{n!}\frac{e^{iu}(u-1)^n}{u}}$

$\textstyle\displaystyle{I=\operatorname{Ei}(iu)+\sum_{n=1}^{\infty}\frac{t_n}{n!}\int\frac{e^{iu}(u-1)^n}{u}du}$

Now there might be a more clever way to get the integral in the sum, but the easiest way I can think of is the Binomial Theorem.

$\textstyle\displaystyle{I=\operatorname{Ei}(iu)+\sum_{n=1}^{\infty}\frac{t_n}{n!}\left(\int\frac{(-1)^n}{u}du+\sum_{k=1}^{n}{n\choose k}(-1)^k\int u^{n-k-1}e^{iu}du\right)}$

Substitute, $v=-iu$ for the integral $\int u^{n-k-1}e^{iu}du$. Then $du=idv$.

$\implies\textstyle\displaystyle{I=\operatorname{Ei}(iu)+\sum_{n=1}^{\infty}\frac{t_n}{n!}\left((-1)^n\ln(u)+\sum_{k=1}^{n}{n\choose k}(-1)^ki^{n-k}\int v^{n-k-1}e^{-v}dv\right)}$

$\implies\textstyle\displaystyle{I=C+\operatorname{Ei}(iu)+\sum_{n=1}^{\infty}\frac{t_n}{n!}\left((-1)^n\ln(u)-\sum_{k=1}^{n}{n\choose k}i^{n+k}\Gamma(n-k,v)\right)}$

$\implies\textstyle\displaystyle{I=C+\operatorname{Ei}(iu)+\sum_{n=1}^{\infty}t_n\left(\frac{(-1)^n\ln(u)}{n!}-\sum_{k=1}^{n}\frac{i^{n+k}\operatorname{Q}(n-k,-iu)}{k!(n-k)}\right)}$

I tried but I can't simplify more than this. And for the $t_n$, I don't know is there closed form for it, I just can't imagine how to even get started to simply $t_n$. So this is it.

Conclusion:-

$\textstyle\displaystyle{\int e^{ix^x}dx=C+\operatorname{Ei}(ix^x)+\sum_{n=1}^{\infty}t_n\left(\frac{(-1)^nx\ln(x)}{n!}+\sum_{k=1}^{n}\frac{i^{n+k}\operatorname{Q}(n-k,-ix^x)}{k!(k-n)}\right)}$

Where,

$\textstyle\displaystyle{t_n=(-1)^n\sum_{k=1}^{n}\left|S_{n}^{(k)}\right|k^k}$

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Sep 23 at 13:39

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