Integration by Parts

The first, tempting, strategy is to use the Taylor series expansion

\begin{equation} \exp(-t^{4}) = 1 - t^{4} + \frac{1}{2}t^{8} +\dots \end{equation}

But the integral doesn't simplify. We just get a sum of divergent integrals:

\begin{equation} I(x) = \int^{\infty}_{x} \left(1 - t^{4} + \frac{1}{2}t^{8} +\dots\right)\,\D t \end{equation}

The integrand may be plotted:

Perhaps surprising how, over the domain \(0\leq x\lt 1\), the integrand is larger than 0.36 and for \(x\lt0.75\) it's nearly 1. We could use a "negative space" argument, treating

\begin{equation} I(x) = I(0) - \displaystyle\int^{x}_{0}\E^{-t^{4}}\,\D t. \end{equation}

We have

\begin{equation} I(0) = \frac{\Gamma(1/4)}{4} \approx 0.90640248 \end{equation}

Believe me? No, you shouldn't. To see this, we do a change of variables to \(\tau=t^{4}\), hence

\begin{equation} \D t = \frac{1}{4}\tau^{-3/4}\,\D\tau \end{equation}

and the integral is

\begin{equation} I(0) = \int^{\infty}_{0}\frac{1}{4}\tau^{-3/4}\E^{-\tau}\,\D\tau \end{equation}

which, looking above at the definition of the gamma function, gives us the desired result.

Now, for the second integral, we can use Taylor expansion to give us

\begin{equation} \begin{split} I_{2}(x) &= \int^{x}_{0}\left(1 - t^{4} + \frac{1}{2}t^{8}+\dots\right)\D t\\ &= x - \frac{x^{5}}{5} + \frac{x^{9}}{18} + \dots \end{split} \end{equation}

Hence we have

\begin{equation} I(x) = \frac{\Gamma(1/4)}{4} - \left[x - \frac{x^{5}}{5} + \frac{x^{9}}{18} + \dots\right]. \end{equation}

This is convergent, but useless for \(x\lesssim 1\). The L2-distance-squared between the three-term approximation and exact integral, over the domain \(0\lt x\lt1\), is about \(4.44854\times 10^{-6}\). The maximum distance is at \(x=1\), which underestimates the true value by \(0.010717\) or so.

The trick for approximating \(I(x)\) for large \(x\) is to deploy integration-by-parts, a classic tool from classical analysis. We observe

\begin{equation} \D(\E^{-t^{4}}) = \E^{-t^{4}}\,\D(-t^{4}) = (-4t^{3})\E^{-t^{4}}\,\D t \end{equation}

hence

\begin{equation} u\,\D v = \frac{-1}{4t^{3}}\D(\E^{-t^{4}}) = \E^{-t^{4}}\,\D t. \end{equation}

Then the integral is

\begin{equation} \begin{split} I(x) &= \int u\,\D v = \left.uv\right|^{t=\infty}_{t=x} - \int v\,\D u\\ &= \frac{\E^{-x^{4}}}{4x^{3}} + \displaystyle\int^{\infty}_{x}\E^{-t^{4}}\left(\frac{3}{4}t^{-t}\right)\D t. \end{split} \end{equation}

The integral in the second line is the remainder term, \(R(x)\). We now claim

\begin{equation} I(x)\sim \frac{\E^{-x^{4}}}{4x^{3}} \end{equation}

with remainder \(R(x)\).

Now, if this is working, if integation by parts is producing the correct asymptotic approximation, then we would have

\begin{equation} I(x)\gg R(x) \end{equation}

and in particular

\begin{equation} \frac{\E^{-x^{4}}}{4x^{3}}\gg R(x). \end{equation}

We'd need to prove it. We will find an upper bound on the remainder term

\begin{equation} |R(x)| = \frac{3}{4}\left|\int^{\infty}_{x}\frac{1}{t^{4}}\E^{-t^{4}}\,\D t\right| \end{equation}

which can be rewritten as

\begin{equation} |R(x)| = \frac{3}{4}\left|\int^{\infty}_{x}\frac{1}{t^{4}}\frac{1}{4t^{4}}\D(\E^{-t^{4}})\right| \end{equation}

Then since \(x^{7}\lt t^{7}\) we have \(x^{-7}\gt t^{-7}\) giving us the bound

\begin{equation} |R(x)| \leq \frac{3}{16x^{7}}\left|\int^{\infty}_{x}\D(\E^{-t^{4}})\right|. \end{equation}

Hence

\begin{equation} |R(x)| \lt\left|\frac{3}{16x^{7}}\E^{-x^{4}}\right| \end{equation}

which is \(x^{-7}\) times smaller than the asymptotic approximation of \(I(x)\). Since this is for large \(x\to\infty\), we find

\begin{equation} 0\leq\lim_{x\to\infty}\frac{|R(x)|}{\E^{-x^{4}}/(4x^{3})}\leq\lim_{x\to\infty}\frac{\frac{3}{16x^{7}}\E^{-x^{4}}}{\E^{-x^{4}}/(4x^{3})}=0. \end{equation}

Hence \(R(x)\ll \E^{-x^{4}}/(4x^{3})\).

The large \(x\) approximation starts working around \(x\approx 3/2\). The small \(x\) approximation works until \(x\approx 1\). The only difficult domain not described by these approximations are \(1\lesssim x\lesssim3/2\).