- #1

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## Main Question or Discussion Point

I have been working on this integral for a long time ,however i still can't solve it because of the comlicated index of exponential.So if any of you have any idear,please tell me ,thank you so much~~

[tex]

\int _{0}^{\infty }\!{{\rm e}^{-2\,{{\it m1}}^{2}-2\,{{\it m2}}^{2}}}{

\pi }^{-1}r \left( 1+{r}^{2} \right) ^{-3}

\left( \left( 1+2\,{{\it m2}}^{2} \right) {r}^{2}+4\,{\it m1}\,{\it

m2}\,br+1+2\,{{\it m1}}^{2} \right) {{\rm e}^{{\frac {2\,{{\it m1}}^{2

}+2\,{{\it m2}}^{2}{r}^{2}+4\,{\it m1}\,{\it m2}\,br}{1+{r}^{2}}}}}{dr}

[/tex]

I have already tried to substitute r by tan(theta),and i have already refer to the handbook <<Table of integrals,series,and products>>,without finding this type of integral.In the end I have refered to <<Analytical calculation of a class of integrals containing exponential and trigonometric functions>>,howerver i find it to represent results in series of Bessel functions ,that is really not what i want.

[tex]

\int _{0}^{\infty }\!{{\rm e}^{-2\,{{\it m1}}^{2}-2\,{{\it m2}}^{2}}}{

\pi }^{-1}r \left( 1+{r}^{2} \right) ^{-3}

\left( \left( 1+2\,{{\it m2}}^{2} \right) {r}^{2}+4\,{\it m1}\,{\it

m2}\,br+1+2\,{{\it m1}}^{2} \right) {{\rm e}^{{\frac {2\,{{\it m1}}^{2

}+2\,{{\it m2}}^{2}{r}^{2}+4\,{\it m1}\,{\it m2}\,br}{1+{r}^{2}}}}}{dr}

[/tex]

I have already tried to substitute r by tan(theta),and i have already refer to the handbook <<Table of integrals,series,and products>>,without finding this type of integral.In the end I have refered to <<Analytical calculation of a class of integrals containing exponential and trigonometric functions>>,howerver i find it to represent results in series of Bessel functions ,that is really not what i want.