**4. Special integrals**

Let us consider the following class of the integrals:

$$\int\_{(0,+\infty)^\prime} f(\varkappa\_1, \varkappa\_2, \dots, \varkappa\_s) d\varkappa\_1 d\varkappa\_2 \dots d\varkappa\_s,\tag{21}$$

We define the behavior of the subintegral function as follows: *f x*ð Þ 1, *x*2, … , *xs* to be [label = ()]

1.close to 1 in the cube 0, ½ � *<sup>a</sup> <sup>s</sup>* as 0 <sup>&</sup>lt;*a*<1;

2.much less than unit as *a*<*xi* < *b*, *b*≥ *a*;

3.equil to zero as *b*≤*xi* < ∞.

Note that very often the integration of functions can be reduced to the linear combination of the integrals similar to Eq. (21) using various replacements of variables.

Below, let us perform a theoretical and numerical analysis how to integrate a model function from our class. The model function is assumed to be *f* � 1 as 0 ≤*xi* ≤*a*, otherwise *<sup>f</sup>* <sup>¼</sup> 0. We take both distribution densities set *<sup>p</sup>*1ð Þ¼ *xi <sup>λ</sup>e*�*λxi* , 0≤*xi* < ∞ and *<sup>p</sup>*2ð Þ¼ *xi* ð Þ *<sup>ω</sup>* <sup>þ</sup> <sup>1</sup> ð Þ <sup>1</sup> � *xi <sup>ω</sup>*, 0<sup>≤</sup> *xi* <sup>&</sup>lt;1 to be examined. Our goal is to determine what of two densities provides the best accuracy of the integral computation with given model function.

If we simulate a random point *<sup>ξ</sup>* <sup>¼</sup> *<sup>ξ</sup>*1, … , *<sup>ξ</sup><sup>s</sup>* ð Þ with densities *<sup>p</sup>*1ð Þ¼ *xi <sup>λ</sup>e*�*λxi* then the integral estimation is given by

$$\begin{aligned} \eta\_s &= \prod\_{i=1}^s \lambda e^{\lambda \xi\_i}, \\ \mathbf{var}\eta\_s &= \mathcal{E}\eta\_s^2 - (\mathcal{E}\eta\_s)^2 = \left[\frac{1}{\lambda} \int\_0^t e^{\lambda r} d\mathbf{x}\right]^s - a^s = \left[\frac{e^{\lambda a} - \mathbf{1}}{\lambda^2}\right]^s - a^s. \end{aligned} \tag{22}$$

Testing the variance **var***η<sup>s</sup>* for the extremum over *λ* we get the minimum condition

$$
\lambda a e^{\lambda a} - 2e^{\lambda a} + 2 = 0.\tag{23}
$$

Let *A* be *λa*; then the equation Eq. (23) is reduced to

$$Ae^A - 2e^A + 2 = 0.\tag{24}$$

The equation above has the unique root at *A* ≈1, 593620. It follows that *λ min* ¼ *A=a*. For such *λ min* the relative error with the 3*σ* rule is given by

$$\frac{3\sigma}{\sqrt{N}a^{\epsilon}} = 3\left(\frac{\left(e^{A}-1\right)^{\epsilon}}{A^{2\epsilon}} - 1\right)^{0.5}/\sqrt{N}.\tag{25}$$

Suppose *<sup>N</sup>* <sup>¼</sup> <sup>9</sup> � <sup>10</sup>6, *<sup>s</sup>* <sup>¼</sup> 10, *<sup>λ</sup>* <sup>¼</sup> *<sup>A</sup>=a*; then the theoretical value of the relative error is approximately 8, 72 � <sup>10</sup>�<sup>3</sup> . The numerical estimation of the relative error is approximately 8, 69 � <sup>10</sup>�<sup>3</sup> as *<sup>a</sup>*∈½ � 0, 1; 0, 001 . Thus, the numerical estimation of one gives a good fit to the predicted value over a wide range of *a*.

Now we discuss the use of the density *<sup>p</sup>*2ð Þ¼ *xi* ð Þ *<sup>ω</sup>* <sup>þ</sup> <sup>1</sup> ð Þ <sup>1</sup> � *xi <sup>ω</sup>*. First, we estimate the second moment of a random value *ηs*:

$$
\mathcal{E}\eta\_s^2 = \left[\frac{\mathbf{1} - (\mathbf{1} - a)^{\mathbf{1} - a}}{\mathbf{1} - a^2}\right]^s. \tag{26}
$$

The parameter *ω* is chosen to be *A=a*; then the expression above is rewritten as follows

$$\mathcal{E}\eta\_s^2 = \left[\frac{1}{1 - A^2/a^2} \left(\mathbf{1} - (\mathbf{1} - a)^{1 - A/a}\right)\right]^\varepsilon. \tag{27}$$

Let us consider E*η*<sup>2</sup> *<sup>s</sup>* as *a* ! 0:

$$\begin{split} \lim\_{a \to 0} \mathcal{E} \eta\_{\iota}^{2} &= \lim\_{a \to 0} \left[ \frac{a^{2}}{a^{2} - A^{2}} \left( \mathbf{1} - (\mathbf{1} - a)^{\mathbf{1} - A/a} \right) \right]' = \\ &= - \lim\_{a \to 0} \left[ \frac{a^{2}}{A^{2}} \left( \mathbf{1} - (\mathbf{1} - a)(\mathbf{1} - a)^{-A/a} \right) \right]' = - \lim\_{a \to 0} \left[ \frac{a^{2}}{A^{2}} \left( \mathbf{1} - (\mathbf{1} - a) \left[ (\mathbf{1} - a)^{-1/a} \right]^{A} \right) \right]' = \\ &= - \lim\_{a \to 0} \left[ \frac{a^{2}}{A^{2}} \left( \mathbf{1} - (\mathbf{1} - a) e^{A} \right) \right]' = - \lim\_{a \to 0} \left[ \frac{a^{2}}{A^{2}} \left( \mathbf{1} - e^{A} \right) \right]' \sim \left[ \frac{\left( e^{A} - \mathbf{1} \right)}{A^{2}} a^{2} \right]'. \end{split} \tag{28}$$

Comparison between Eqs. (22) and (28) allows to make the following conclusion. If *w* is chosen to be *A=a* then the asymptotics of variances, as *a* ! 0, are the same in the densities set of *p*1ð Þ *xi* , *p*2ð Þ *xi* . In the numerical simulation the relative accuracy of <sup>≈</sup> 8, 68 � <sup>10</sup>�<sup>3</sup> is reached as *<sup>N</sup>* <sup>¼</sup> <sup>9</sup> � 106, *<sup>s</sup>* <sup>¼</sup> 10,*<sup>ω</sup>* <sup>¼</sup> *<sup>A</sup>=a*, *<sup>a</sup>*<sup>∈</sup> ½ � 0, 01; 0, 001 .

Let us turn now to the integral

$$I(f) = \int\_{[0,1]^{\mathbb{N}}} f(\mathbf{x}\_1, \mathbf{x}\_2, \dots, \mathbf{x}\_{10}) d\mathbf{x}\_1 d\mathbf{x}\_2 \dots d\mathbf{x}\_{10},\tag{29}$$

where

$$f(\mathbf{x}\_1, \mathbf{x}\_2, \dots, \mathbf{x}\_{10}) = \begin{cases} 1, & \mathbf{0} \le \mathbf{x}\_i \le \mathbf{1}/4, \\ \mathbf{0}, & \text{otherwise.} \end{cases} \tag{30}$$

We put *p x*ð Þ� 1; then *I f*ð Þ¼ *I f*ð Þ¼ *<sup>=</sup><sup>p</sup> I f* <sup>2</sup> *<sup>=</sup><sup>p</sup>* � � <sup>¼</sup> ð Þ <sup>1</sup>*=*<sup>4</sup> <sup>10</sup> <sup>≈</sup>9, 5367 � <sup>10</sup>�7*:* For *N* ¼ 640000 all realizations are turned out to be equal to zero, i.e., *q*640000ð Þ¼ *x* 0 as 0≤*xi* ≤1*=*4. In this case, we have

$$\int\_{[0,1]^{10}} p(\mathbf{x}) \frac{f(\mathbf{x})}{p(\mathbf{x})} d\mathbf{x} - \int\_{[0,1]^{10}} q\_{640000} \frac{f(\mathbf{x})}{p(\mathbf{x})} d\mathbf{x} = \left(\mathbf{1}/4\right)^{10} - \mathbf{0} = \left(\mathbf{1}/4\right)^{10}.\tag{31}$$

In accordance with *N*, the realizations numbers of *η<sup>i</sup>* are turned out to be equal to 1 as *N* ¼ 810000; equal to 3 as *N* ¼ 4000000; equal to 15 as *N* ¼ 16000000.

We now take *p x*ð Þ¼*<sup>i</sup>* ð Þ *<sup>ω</sup>* <sup>þ</sup> <sup>1</sup> ð Þ <sup>1</sup> � *xi <sup>ω</sup>* in the unit cube 0, 1 ½ �<sup>10</sup> ð Þ *<sup>ω</sup>*><sup>1</sup> . Such choice provides the gross realizations of points in 0, 1 ½ � *<sup>=</sup>*<sup>4</sup> <sup>10</sup> and as consequence, we get benefit in quality of random values (simultaneously, we have decrease of the estimation variance, and as consequence decrease of the statistical error with the 3*σ* rule.) **Table 5** shows the calculations results for *N* ¼ 9000000. Note that *σ*^ reaches the minimum as *<sup>ω</sup>* <sup>¼</sup> 5. In this case, we have *p x* ^ð Þ¼ 6 1ð Þ � *<sup>x</sup>* <sup>5</sup> , *I f*ð Þ¼ *<sup>=</sup>p*^ *I f* <sup>2</sup> *<sup>=</sup>p*^ � �≈3, 49 � <sup>10</sup>�<sup>11</sup>*:* Making the more detailed research for both *<sup>ω</sup>* <sup>¼</sup> 5 and the theoretical value *<sup>σ</sup>* <sup>≈</sup>5, 83 � <sup>10</sup>�<sup>6</sup> we get the results represented in **Table 6**. If the function


**Table 5.**

*The results of numerical calculations for the integral Eq. (29) at various ω values.*


**Table 6.**

*The results of numerical calculations for the integral Eq. (29) at ω* ¼ 5*.*

*f x*ð Þ 1, *<sup>x</sup>*2, … , *<sup>x</sup>*<sup>10</sup> close to some constant in 0, 1 ½ � *<sup>=</sup>*<sup>4</sup> <sup>10</sup> and small out of this interval then we can advise to use *<sup>p</sup>*^ <sup>¼</sup> 6 1ð Þ � *<sup>x</sup>* <sup>5</sup> to calculate the integral in 0, 1 ½ �10.
