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Carlo Beenakker
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From the comments I understand that the OP seeks an approximation of $$I=\int_{x_1}^{x_2} x^{-a} \text{erf}\left( b - c x^{-d} \right)\, dx$$ for $x_2\gg x_1\gg 1$. A complication which will limit the accuracy of the approximation is that $d\ll 1$. If I ignore that for a moment, and assume all coefficients $a,b,c,d$ are of order unity, then a large-$x$ expansion of the integrand gives the approximation $$I_{\text{appr}}={x_2}^{-a} {x_1}^{-a} \left(\frac{2 e^{-b^2} c {x_2}^{-d} {x_1}^{-d} \left({x_2} {x_1}^{a+d}-{x_1} {x_2}^{a+d}\right)}{\sqrt{\pi } (a+d-1)}+\frac{\text{erf}(b) \left({x_1} {x_2}^a-{x_2} {x_1}^a\right)}{a-1}\right).$$$$I_{\text{appr}}=\int_{x_1}^{x_2}x^{-a}\left(\text{erf}(b)-\frac{2c e^{-b^2} }{x^{d}\sqrt{\pi }}\right)\,dx$$ $$\qquad={x_2}^{-a} {x_1}^{-a} \left(\frac{2 e^{-b^2} c {x_2}^{-d} {x_1}^{-d} \left({x_2} {x_1}^{a+d}-{x_1} {x_2}^{a+d}\right)}{\sqrt{\pi } (a+d-1)}+\frac{\text{erf}(b) \left({x_1} {x_2}^a-{x_2} {x_1}^a\right)}{a-1}\right).$$ The values of interest to the OP are $\{a,b,c,d,x_1,x_2\}=\{1.2743, 16.33, 18.7525, 0.0308, 284.959, 712090\}$ In this case $I=0.621072$ while $I_{\text{appr}}=0.682988$, an error of 10%. If the parameter $d$ is increased slightly to $0.05$ the agreement improves to four decimal places.

From the comments I understand that the OP seeks an approximation of $$I=\int_{x_1}^{x_2} x^{-a} \text{erf}\left( b - c x^{-d} \right)\, dx$$ for $x_2\gg x_1\gg 1$. A complication which will limit the accuracy of the approximation is that $d\ll 1$. If I ignore that for a moment, and assume all coefficients $a,b,c,d$ are of order unity, then a large-$x$ expansion of the integrand gives the approximation $$I_{\text{appr}}={x_2}^{-a} {x_1}^{-a} \left(\frac{2 e^{-b^2} c {x_2}^{-d} {x_1}^{-d} \left({x_2} {x_1}^{a+d}-{x_1} {x_2}^{a+d}\right)}{\sqrt{\pi } (a+d-1)}+\frac{\text{erf}(b) \left({x_1} {x_2}^a-{x_2} {x_1}^a\right)}{a-1}\right).$$ The values of interest to the OP are $\{a,b,c,d,x_1,x_2\}=\{1.2743, 16.33, 18.7525, 0.0308, 284.959, 712090\}$ In this case $I=0.621072$ while $I_{\text{appr}}=0.682988$, an error of 10%. If the parameter $d$ is increased slightly to $0.05$ the agreement improves to four decimal places.

From the comments I understand that the OP seeks an approximation of $$I=\int_{x_1}^{x_2} x^{-a} \text{erf}\left( b - c x^{-d} \right)\, dx$$ for $x_2\gg x_1\gg 1$. A complication which will limit the accuracy of the approximation is that $d\ll 1$. If I ignore that for a moment, and assume all coefficients $a,b,c,d$ are of order unity, then a large-$x$ expansion of the integrand gives the approximation $$I_{\text{appr}}=\int_{x_1}^{x_2}x^{-a}\left(\text{erf}(b)-\frac{2c e^{-b^2} }{x^{d}\sqrt{\pi }}\right)\,dx$$ $$\qquad={x_2}^{-a} {x_1}^{-a} \left(\frac{2 e^{-b^2} c {x_2}^{-d} {x_1}^{-d} \left({x_2} {x_1}^{a+d}-{x_1} {x_2}^{a+d}\right)}{\sqrt{\pi } (a+d-1)}+\frac{\text{erf}(b) \left({x_1} {x_2}^a-{x_2} {x_1}^a\right)}{a-1}\right).$$ The values of interest to the OP are $\{a,b,c,d,x_1,x_2\}=\{1.2743, 16.33, 18.7525, 0.0308, 284.959, 712090\}$ In this case $I=0.621072$ while $I_{\text{appr}}=0.682988$, an error of 10%. If the parameter $d$ is increased slightly to $0.05$ the agreement improves to four decimal places.

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Carlo Beenakker
  • 188.1k
  • 18
  • 448
  • 651

From the comments I understand that the OP seeks an approximation of $$I=\int_{x_1}^{x_2} x^{-a} \text{erf}\left( b - c x^{-d} \right)\, dx$$ for $x_2\gg x_1\gg 1$. A complication which will limit the accuracy of the approximation is that $d\ll 1$. If I ignore that for a moment, and assume all coefficients $a,b,c,d$ are of order unity, then a large-$x$ expansion of the integrand gives the approximation $$I_{\text{appr}}={x_2}^{-a} {x_1}^{-a} \left(\frac{2 e^{-b^2} c {x_2}^{-d} {x_1}^{-d} \left({x_2} {x_1}^{a+d}-{x_1} {x_2}^{a+d}\right)}{\sqrt{\pi } (a+d-1)}+\frac{\text{erf}(b) \left({x_1} {x_2}^a-{x_2} {x_1}^a\right)}{a-1}\right).$$ The values of interest to the OP are $\{a,b,c,d,x_1,x_2\}=\{1.2743, 16.33, 18.7525, 0.0308, 284.959, 712090\}$ In this case $I=0.621072$ while $I_{\text{appr}}=0.682988$, an error of 10%. If the parameter $d$ is increased slightly to $0.05$ the agreement improves to four decimal places.