# Why does this integral not converge?

I have a rather simple integral that I want to solve in Mathematica.

Integrate[Sin[x*a/2]^4/x^1, {x, 0, Infinity}]

where a is a real parameter larger than zero.

Upon putting this in Mathematica, it says the integral doesn't converge. Are there any tricks that I can use to make this converge, or can I trust Mathematica that it doesn't converge?

• It is of the form 1/x, so what makes you think it does converge? Commented Feb 5 at 9:52
• Well cause sin(x)/x is one for x to zero Commented Feb 5 at 9:53
• But on the other end (x infinity) it doesn't… You can do 1 till infinity and it still does not converge. Looking at the far end is enough in this case… Commented Feb 5 at 9:55
• It would converge if the exponent of x in denominator is bigger than 1. Commented Feb 5 at 9:56
• This is an analysis question as much as a Mathematica question. Yes, it diverges. One can bound it from below by a divergent sum. Commented Feb 5 at 14:51

You can first do the integral symbolically:

Integrate[Sin[x*1/2]^4/x^1, x]


which gives you:

1/8 (-4 CosIntegral[x] + CosIntegral[2 x] + 3 Log[x/2])


As you can see by filling in the bounds, the Log prevents it from being a finite value…

If the sine function were not taken to an even power, its alternation of sign would give a convergent (though not absolutely convergent) integral. But, with that even power, the parts of the integral where $$(\sin x)^4$$ is $$\ge 1/2$$ gives a lower bound of the whole by a definitely divergent integral.

On another hand, if/when one really needs to make sense of "divergent integrals", especially in inner steps of a proof/computation, sometimes it does suffice to view the thing as a Fourier transform of a tempered distribution... which not only "makes some sense", but is compatible with more classical styles of computation of integrals and such.