3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $$ \phi := \frac{c}{\sqrt{c^2+\alpha^2}} \,. $$

$\endgroup$
  • $\begingroup$ 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

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.