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I have to understand how to put the parameter $a$ when it comes to generalise the function under the integral sign.

For istance I tried to solve this integral

$$\int_{0}^{\infty}{\frac{\sin(x)}{x^2+1}}dx$$

But I didn't know how to generalise the integrand.

Are there specific rules for certain groups of functions or should I just have a good abstraction sense?

us er
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  • Thanks to whomever edited my question – us er Aug 16 '20 at 13:04
  • The integral you've given doesn't have a nice closed form according to Mathematica. Did you have a reason to expect a nice answer? – Patrick Stevens Aug 16 '20 at 13:10
  • The integral is definite, there are many integrals which do not have closed-form primitives but when integrated with integration extremes have an answer. The integral sine for example, from 0 to infinity is pi/2 – us er Aug 16 '20 at 13:13
  • Yes, but this particular integral does not in fact have a closed form, according to Mathematica. The mere fact that some integrals do have closed forms does not imply that this one does. – Patrick Stevens Aug 16 '20 at 13:25

1 Answers1

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Practicing and trying different things helps most when it comes to integration techniques. In many cases I don't think spotting Feymann's technique is easy unless you've seen it done before. The only general tip I can think of is to try finding things which are easy to differentiate. In the given example, it is sensible to consider

$$\int_0^\infty\frac{\sin(x)e^{ax}}{x^2+1}~\mathrm dx,\qquad\int_0^\infty\frac{\sin(ax)}{x^2+1}~\mathrm dx,\qquad\int_0^\infty\frac{\sin(x)}{x^2+a}~\mathrm dx$$

but it is usually less sensible to consider

$$\int_0^\infty\frac{\sin(x)}{x^a+1}~\mathrm dx$$

Incidentally, the first two parameterizations allow your integral to be solved, the third may allow generalizations to denominators of the form $(x^2+a)^n$, and the last doesn't seem very helpful.

Simply Beautiful Art
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