0
$\begingroup$
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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user64494 Jun 19 '19 at 4:32
  • $\begingroup$ my c is a variable of an another expression $\endgroup$ – user105697 Jun 19 '19 at 4:55
  • $\begingroup$ So what? I don't see a problem. $\endgroup$ – user64494 Jun 19 '19 at 5:00
  • $\begingroup$ you can't just put a numerical value to c. It will be used as a variable in further calculation $\endgroup$ – user105697 Jun 19 '19 at 5:03
2
$\begingroup$

If you insist on "symbolic expression", then the following does the job.

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], 15], {x, Rationalize[5.03], Rationalize[994.97]}]

$$ i \left(-\frac{1}{3} 80 \sqrt{10 \pi } c^{3/2} \text{erf}\left(\frac{\sqrt{c}}{5 \sqrt{2}}\right)+\frac{8}{15} \sqrt{\frac{2 \pi }{5}} (c+125) c^{3/2} \text{erf}\left(10 \sqrt{\frac{10}{503}} \sqrt{c}\right)+\frac{8}{15} \sqrt{\frac{2 \pi }{5}} (c+125) c^{3/2} \text{erf}\left(10 \sqrt{\frac{10}{99497}} \sqrt{c}\right)-\frac{16}{15} \sqrt{\frac{2 \pi }{5}} c^{5/2} \text{erf}\left(\frac{\sqrt{c}}{5 \sqrt{2}}\right)+\frac{4}{375} \sqrt{503} e^{-\frac{1000 c}{503}} c^2+\frac{4}{375} \sqrt{99497} e^{-\frac{1000 c}{99497}} c^2-\frac{32 e^{-\frac{c}{50}} c^2}{3 \sqrt{5}}-\frac{640}{3} \sqrt{5} e^{-\frac{c}{50}} c+\frac{249497 \sqrt{503} e^{-\frac{1000 c}{503}} c}{187500}+\frac{150503 \sqrt{99497} e^{-\frac{1000 c}{99497}} c}{187500}+\frac{8000}{3} \sqrt{5} e^{-\frac{c}{50}}+\frac{4824709027 \sqrt{99497} e^{-\frac{1000 c}{99497}}}{375000000}-\frac{124990973 \sqrt{503} e^{-\frac{1000 c}{503}}}{375000000}\right)$$

$\endgroup$
  • $\begingroup$ what this 15 inside the integral means $\endgroup$ – user105697 Jun 19 '19 at 6:04
  • $\begingroup$ @user105697: This is the same as Rationalize[..,10^(-15)]. See the help to Rationalize for details. $\endgroup$ – user64494 Jun 19 '19 at 6:19
  • $\begingroup$ ok you are right but i actually stuck doing this : LogLogPlot[ 1.1183*3.086*10^-12 c^(1/2)*2.598*10^-3 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], 15], {x, Rationalize[5.03], Rationalize[994.97]}], {c, 10^0, 10^6}, PlotRange -> {{10^0, 10^4}, {10^-18, 10^-8}}] and it is giving a blank graph $\endgroup$ – user105697 Jun 19 '19 at 6:29
  • $\begingroup$ 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 $$\int_{\frac{503}{100}}^{\frac{99497}{100}} \sqrt{\frac{21272654}{21707411}-\left(1-\frac{x}{500}\right)^2} \sqrt{x} e^{-\frac{107 c}{10 x}} \, dx .$$ $\endgroup$ – user64494 Jun 19 '19 at 7:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.