This emerged while I was investigating this question, i.e. the solution to the definite integral $$I_x = \int_0^\infty\left(5x^5+x\right)\operatorname{erfc}\left(x^5+x\right)\,dx$$

In a comment, its value is given as 0.2119539... using the Maple solver, and I have verified this value by using another on-line definite integral calculator.

But manipulating the integral leads me to a different solution. Specifically,

define $y=h(x) = x^5 +x$, and we have $dy = h'dx$, with $h' = 5x^4+1$. Also, $h(x)$ admits an inverse, so $x = h^{-1}(y)$. Then

$$\int_0^\infty\left(5x^5+x\right)\operatorname{erfc}\left(x^5+x\right)\,dx = \int_0^\infty x\left(5x^4+1\right)\operatorname{erfc}\left(x^5+x\right)\,dx$$ $$= \int_0^\infty xh'\operatorname{erfc}\left(y\right)\,dx = \int_0^\infty h^{-1}(y)\operatorname{erfc}\left(y\right)\,dy$$

We also have $\operatorname{erfc}(y) = 1-\operatorname{erf}(y) = 1- \left(2\Phi(\sqrt 2 y)-1\right) \Rightarrow \operatorname{erfc}(y) = 2\Phi(-\sqrt 2 y)$ where $\Phi()$ is the standard normal cdf.

Then $$\int_0^\infty h^{-1}(y)\operatorname{erfc}\left(y\right)\,dy = \int_0^\infty h^{-1}(y)2\Phi(-\sqrt 2 y)\,dy = 2\int_0^\infty \int_{-\infty}^{-\sqrt 2 y}h^{-1}(y)\phi(t)\,dtdy$$ where $\phi()$ is the standard normal pdf. Interchange the limits of integration from

\begin{matrix} 0 \le y \le \infty\\ -\infty \le t \le - \sqrt 2 y \end{matrix}

to \begin{matrix} -\infty \le t \le 0\\ 0 \le y \le - t/\sqrt 2 \end{matrix} to obtain

$$I_x = 2\int_{-\infty}^0\phi(t) \int_{0}^{- t/\sqrt 2}h^{-1}(y)\,dydt$$

using inverse function integration we have (HERE COMES THE MISTAKE)

$$\int_{0}^{- t/\sqrt 2}h^{-1}(y)dy = xh(x)\Big |_0^{- t/\sqrt 2} - \int_{0}^{- t/\sqrt 2}h(x)dx $$ $$= (x^6 + x^2)\Big|_0^{- t/\sqrt 2} - \left((\frac 16 x^6 + \frac 12 x^2)\Big|_0^{- t/\sqrt 2}\right) = \left(\frac 56 x^6 + \frac 12 x^2\right)\Big|_0^{- t/\sqrt 2} = \frac {5}{48} t^6 + \frac 14 t^2$$

The correct expression for this case is $$\int_{0}^{- t/\sqrt 2}h^{-1}(y)dy = xh(x)\Big |_0^{h^{-1}(- t/\sqrt 2)} - \int_{0}^{h^{-1}(- t/\sqrt 2)}h(x)dx $$

Inserting into the integral we have

$$I_x = \frac {5}{48}\cdot 2\int_{-\infty}^0t^6\phi(t)dt + \frac {1}{4}\cdot 2\int_{-\infty}^0t^2\phi(t)dt$$

The integrals now represent even half-moments of the standard normal distribution, which are symmetric around zero and since they are multiplied by $2$ they equal whole moments. So $$I_x = \frac {5}{48}E_{\Phi}(t^6) + \frac {1}{4}E_{\Phi}(t^2) = \frac {5}{48}(3\cdot 5) + \frac {1}{4} \cdot 1 = \frac{25}{16} + \frac{4}{16} = \frac {29}{16} = 1.8125$$

Either there is a silly calculating mistake somewhere, or at some point I must be doing something illegal (or both).

I would greatly appreciate if somebody would point these to me, or just give me a hint.


You misapplied the inverse function integration. The identity is

$$\int f(x)\,dx = x\cdot f(x) - G(f(x)),$$

where $G$ is the indefinite integral of $f^{-1}$.

With $f = h^{-1}$, and $H(x) = \int h(t)\,dt = \frac16 x^6 + \frac12 x^2$, you get

$$\begin{align} \int_0^{-t/\sqrt{2}} h^{-1}(y)\,dy &= yh^{-1}(y)\Big\lvert_0^{-t/\sqrt{2}} - H(h^{-1}(y))\Big\lvert_0^{-t/\sqrt{2}}\\ &= -\frac{t}{\sqrt{2}}h^{-1}\left(-\frac{t}{\sqrt{2}}\right) - H\left(h^{-1}\left(-\frac{t}{\sqrt{2}}\right)\right), \end{align}$$

which I don't see yielding a nice expression.

In response to your comment, when you apply the inverse function integration with $f = h$, you still have - with $H' = h$ and $K' = h^{-1}$

$$H(x) = x\cdot h(x) - K(h(x)),$$

so in order to evaluate

$$\int_u^v h^{-1}(y)\,dy = K(v) - K(u)$$

you still must apply $h^{-1}$ to the bounds to obtain the values to plug into $x\cdot h(x)$ and $H$. That leads (unsurprisingly) to the same expression as above.

  • $\begingroup$ Thank you for your answer, I am still trying to figure this one out. However, I am applying the inverse function integration using $ f = h$. Then what I have is $G(f(x)) = G(y)$ and so I look at $G(f(x)) = G(y) = \int f^{-1}(y)\,dy = x\cdot f(x) - \int f(x)\,dx$. Is this not correct? (I have also corrected notation in the question to make this clear). One more try please! $\endgroup$ – Alecos Papadopoulos Jan 3 '14 at 15:39
  • $\begingroup$ But then your $x\cdot f(x)$ becomes $h^{-1}(y)\cdot y$, and you also must apply $h^{-1}$ to the limits of $\int h$. It leads to the same expression. $\endgroup$ – Daniel Fischer Jan 3 '14 at 16:04
  • $\begingroup$ Thank you. After my comment I have started to suspect that it has to be a matter of "internal consistency" and integration limits, in the application of inverse function integration -and now I see it. $\endgroup$ – Alecos Papadopoulos Jan 3 '14 at 17:12

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