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Mathematics Mathematica Algebra Calculus Wolfram Language Symbolic Computations

Compute the following integral?

Jose richard

Posted 6 years ago

I let Mathematica run overnight the following code Integrate[(r^2 Cos[A r Cos[x]] Exp[-r^2/2])/(a^2 Cos[x]^2 + b^2 Sin[x]^2), {r, 0, Infinity}, {x, 0, Pi/2}] but it was unable to compute the closed form in that time. However, Mathematica also didnt give up. Do you think Mathematica can do it?

POSTED BY: Jose richard

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Dent de Lion

Dent de Lion, Privatier

Posted 6 years ago

Doing the `r` under assumptions is the first step In[2]:= Integrate[(r^2 Cos[A r Cos[x]] Exp[-r^2/2])/(a^2 Cos[x]^2 + b^2 Sin[x]^2), {r, 0, Infinity},(* {x,0,Pi/2} *) Assumptions -> {A > 0, a > 0, b > 0}] Out[2]= (E^(-(1/2) A^2 Cos[x]^2) Sqrt[\[Pi]/2] (1 - A^2 Cos[x]^2))/( a^2 Cos[x]^2 + b^2 Sin[x]^2) now look for a meaningful substitution, you need $0 \leq x \leq \frac{\pi}{2}$. And as a preliminary step try $a = b$ to get rid of the denominator.

Doing the r under assumptions is the first step

In[2]:= Integrate[(r^2 Cos[A r Cos[x]] Exp[-r^2/2])/(a^2 Cos[x]^2 + 
    b^2 Sin[x]^2), {r, 0, Infinity},(* {x,0,Pi/2} *) 
 Assumptions -> {A > 0, a > 0, b > 0}]

Out[2]= (E^(-(1/2) A^2 Cos[x]^2) Sqrt[\[Pi]/2] (1 - A^2 Cos[x]^2))/(
a^2 Cos[x]^2 + b^2 Sin[x]^2)

now look for a meaningful substitution, you need $0 \leq x \leq \frac{\pi}{2}$. And as a preliminary step try $a = b$ to get rid of the denominator.

POSTED BY: Dent de Lion

Joshua Champion

Joshua Champion, Waiter

Posted 6 years ago

Another thing to point out, is that there isn't always an answer to an integral, as my teacher best put it "99% of all integrals don't have antiderivatives." So try making those constant coefficients into variables and write the integral as a function of 3 variables (assuming those three constants are A, a, and b) so that you can work around that instead. You can also plot the newly defined function with respect to two variables and change the third one each time you want to run it. That's my best suggestion. If you want to know more, just ask.