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$$\int_{-\infty}^{\infty} \frac{e^{-\alpha x^2}}{(x^2+c^2)(\sqrt{x^2+u^2})} dx$$

I tried to do it like this:

$$\int_{-\infty}^{\infty} \frac{e^{-\alpha x^2}}{(x^2+c^2)(\sqrt{x^2+u^2})} dx=\int_{-\infty}^{\infty} \frac{e^{-\alpha x^2}\sqrt{x^2+u^2}}{(x^2+c^2)(x^2+u^2)} dx$$ but

$$\frac{1}{x^2+c^2}=\int_{0}^{\infty}e^{-k(x^2+c^2)} dk $$ and $$\frac{1}{x^2+u^2}=\int_{0}^{\infty}e^{-k'(x^2+u^2)} dk' $$ so

$$\int_{-\infty}^{\infty} \frac{e^{-\alpha x^2}}{(x^2+c^2)(\sqrt{x^2+u^2})} dx= \int_{-\infty}^{\infty} e^{-(\alpha+k+k')x^2}\sqrt{x^2+u^2}dx\int_{0}^{\infty} e^{-kc^2}dk... $$

How to solve this first integral?

  • 2
    One parameter among $(\alpha,c,u)$ can be removed through a suitable substitution of the form $x=\kappa z, dx=\kappa\,dz$. Then you may switch to Fourier transforms and that gives a non-elementary integral related with the Bessel K function and the error function. – 2017-01-12
  • 1
    In general, the integral of a continuous, even, positive, integrable function cannot be zero. – 2017-01-12
  • 0
    This [seems to be a duplicate](http://math.stackexchange.com/questions/2093579/how-can-i-solve-this) – 2017-01-13

0 Answers 0