# Is there any way to do this integral without expanding the exponential power?

Integrate[
x^(1/2) Exp[-10.7*
c/x] Sqrt[(1 - (125/1000)^2) (1 - (2*4.18/125)^2) - (1 - (2 x/
1000))^2], {x, 5.03, 994.97}]

or can i do this indefinite integral without the numerical limits. My input is unable to do it

• Try like NIntegrate[ x^(1/2) Exp[-10.7* c/x] Sqrt[(1 - (125/1000)^2) (1 - (2*4.18/125)^2) - (1 - (2 x/ 1000))^2] /. c -> 2.1, {x, 5.03, 994.97}] which produces 15686.7. – user64494 Jun 19 at 4:32
• my c is a variable of an another expression – user105697 Jun 19 at 4:55
• So what? I don't see a problem. – user64494 Jun 19 at 5:00
• you can't just put a numerical value to c. It will be used as a variable in further calculation – user105697 Jun 19 at 5:03

$$i \left(-\frac{1}{3} 80 \sqrt{10 \pi } c^{3/2} \text{erf}\left(\frac{\sqrt{c}}{5 \sqrt{2}}\right)+\frac{8}{15} \sqrt{\frac{2 \pi }{5}} (c+125) c^{3/2} \text{erf}\left(10 \sqrt{\frac{10}{503}} \sqrt{c}\right)+\frac{8}{15} \sqrt{\frac{2 \pi }{5}} (c+125) c^{3/2} \text{erf}\left(10 \sqrt{\frac{10}{99497}} \sqrt{c}\right)-\frac{16}{15} \sqrt{\frac{2 \pi }{5}} c^{5/2} \text{erf}\left(\frac{\sqrt{c}}{5 \sqrt{2}}\right)+\frac{4}{375} \sqrt{503} e^{-\frac{1000 c}{503}} c^2+\frac{4}{375} \sqrt{99497} e^{-\frac{1000 c}{99497}} c^2-\frac{32 e^{-\frac{c}{50}} c^2}{3 \sqrt{5}}-\frac{640}{3} \sqrt{5} e^{-\frac{c}{50}} c+\frac{249497 \sqrt{503} e^{-\frac{1000 c}{503}} c}{187500}+\frac{150503 \sqrt{99497} e^{-\frac{1000 c}{99497}} c}{187500}+\frac{8000}{3} \sqrt{5} e^{-\frac{c}{50}}+\frac{4824709027 \sqrt{99497} e^{-\frac{1000 c}{99497}}}{375000000}-\frac{124990973 \sqrt{503} e^{-\frac{1000 c}{503}}}{375000000}\right)$$
• Sorry, my answer is doubtful in view of Integrate[ x^(1/2) Rationalize[ Exp[-10.7* c/x] Sqrt[(1 - (125/1000)^2)*(1 - (2*4.18/125)^2) - (1 - (2 x/ 1000))^2], 10^-15], {x, Rationalize[5.03], Rationalize[994.97]}] which performs $$\int_{\frac{503}{100}}^{\frac{99497}{100}} \sqrt{\frac{21272654}{21707411}-\left(1-\frac{x}{500}\right)^2} \sqrt{x} e^{-\frac{107 c}{10 x}} \, dx .$$ – user64494 Jun 19 at 7:18