I want to evaluate analytically the integral $\int _ {-1} ^ {1} \mathrm{d} x \frac{1}{ax+b} e^{cx} \sqrt{1-x^2}$, where $a$, $b$ and $c$ are real numbers.

I tried Mathematica, but with no success. Any ideas?

  • $\begingroup$ This isn't worth being an answer, but I would start by rewriting a bit and differentating in the exponent to get rid of the fraction. $\endgroup$ – user111187 May 12 '14 at 20:45
  • $\begingroup$ Substitution x=sinϕ may also be useful. Then the integrand can be expressed as a function of sinϕ only. $\endgroup$ – Urgje May 12 '14 at 21:15
  • $\begingroup$ Could user111187 be a bit more specific? $\endgroup$ – user127054 May 12 '14 at 21:21
  • $\begingroup$ @user127054 My idea was to write the integrand as $\frac{1}{a} \frac{dx}{x+b/a} e^{c(x+b/a)-bc/a}$, take the factor $e^{-bc/a}$ out of the integral, and then differentiate with respect to $c$ in order to remove the $\frac{1}{x+b/a}$ factor. The problem is that you need to integrate back in order to get the original integral. $\endgroup$ – user111187 May 13 '14 at 6:50

If the quantity $$ \beta(x) := \frac{\sqrt{1-x^2}}{ax + b} $$ is such that $\beta(x) \in [-1,1]$ you can define $$ \sin(\alpha x) = \beta \,. $$ The integral then reduces to $$ I = \int_{-1}^1 e^{cx}\,\sin(\alpha x)\,{\rm d}x \,. $$ From Wikipedia, $$ I = \left.\frac{e^{cx}}{\sqrt{c^2+\alpha^2}}\,\sin(\alpha x - \phi)\right|_{-1}^1 $$ where $$ \phi := \frac{c}{\sqrt{c^2+\alpha^2}} \,. $$

  • $\begingroup$ What is $\alpha$? $\endgroup$ – user111187 May 13 '14 at 6:45
  • $\begingroup$ I was thinking of expressing $\beta(x)$ in terms of a Fourier series expansion in sines and cosines. I'll try to correct the answer and make it more explicit when I get some free time. $\endgroup$ – Biswajit Banerjee May 13 '14 at 22:16

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