# Re: tricky integral

## Main Question or Discussion Point

$$\int \frac{3}{2}x^{-\frac{1}{2}}\cdot e^{\frac{3}{2}x^{-2}}\,dx$$

I saw this integral elsewhere and apparently it's insoluble in elementary functions. Anyone got any ideas how to tackle this one, other than as a Taylor series? I couldn't get anywhere with integration by parts, any substitutions that might provide a solution, imaginary or some sort of unusual function? Just curious to see how you would do this.

Just for fun in case anyone's bored.

It comes to something when you're so bored you go trawling for integrals you can't solve.

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What about the substitution u = sqrt(x)....? Then you have dx/sqrt(x) = du, which cancels the first factor so you're left with the integral over $e^{u^{-4}}$

Uhm....yes

I'm just confused now, mathematica says it is impossible to solve this. I tried by parts and came a bit stuck and yet. This either brings up the idea that I am in fact typing it into calc101 wrong, which :

This is my entry.

3/2*x^(-1/2)*e^(3/2x)^(-2)

And the answer is sorry calc101 can't do this.

And then.

The surprisingly easy answer pops out.

$$\frac{3}{2x^{-\frac{1}{2}}} \cdot e^{\frac{3}{2} x^{\frac{1}{2}}$$

So I'm lost now?

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Some more brackets in the exponent, maybe?

3/2*x^(-1/2)*e^((3/2)*x^(-2))

Mathematica gives me a solution ... not as compact as yours though ...

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Some more brackets in the exponent, maybe?

3/2*x^(-1/2)*e^((3/2x)^(-2))

Mathematica gives me a solution ... not as compact as yours though ...

It obviously wasn't as hard as I thought. Thanks Pete.

Actually even with the extra brackets calc101 has a problem. That's really odd? It doesn't usually fall down on easy integrals?

EDIT: I get a gamma function with Mathematica. Hehe.

Anyway if anyone would care to show how to do this integral beyond the substitution given above I'd be grateful.

I get the impression this isn't as straightforward as it appears? It might be interesting to see how mathematica gets such a tortuous answer.

http://www.freewebs.com/mypicturesandsht/mathematica.gif"

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Ok rather than start another thread here's another one I can't get, actually from this forum, but as its marked solved, I doubt anyone will mind?

ok

$\int \sin^{\frac{3}{2}} (x)\;dx\rightarrow \int\sqrt{sin^3(x)}\;dx=?$

Anyone know of a substitution here that can solve this. I can't see anything obvious.

I tried solving it on my maths program but it just spat out the original answer, which either means it's unsolvable in elementary functions or it's broken. If that's the case can anyone go about showing how we get a non explicit answer, just for form.

Mathematica gives an answer but it involves an http://mathworld.wolfram.com/EllipticIntegraloftheFirstKind.html" [Broken] which is lost on me. Anyone explain what they mean, or can anyone derive an answer normally?

Just idle curiosity really.

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Hootenanny
Staff Emeritus
Gold Member
Hi SD I think I have it in terms of elementary functions (unless I've made a daft mistake somewhere).

$$\int\sqrt{\sin^3x}\;\; dx = \int \sin x\sqrt{\sin x} \;\;dx$$

We integrate by parts with,

$$u = \sqrt{\sin x} \Rightarrow \frac{du}{dx} = \frac{\cos x}{2\sqrt{\sin x}}$$

$$dv = \sin x \Rightarrow v = -\cos x$$

Hence,

$$\int \sqrt{\sin^3 x} dx = -\cos x\sqrt{\sin x} - \frac{1}{2}\int\frac{\cos x}{\sqrt{\sin x}} dx$$

We now make a substitution,

$$t = \sin x \Rightarrow \frac{dt}{dx} = \cos x$$

And the integral becomes,

$$\int \sqrt{\sin^3 x} dx = -\cos x\sqrt{\sin x} - \frac{1}{2}\int\frac{dt}{\sqrt{t}}$$

$$= -\cos x\sqrt{\sin x} - \frac{1}{2}\left(-\frac{t^{-3/2}}{2}\right) + \text{ const.}$$

$$= -\cos x\sqrt{\sin x} + \frac{1}{4}\left(\sin^{-3/2}\right) + \text{ const.}$$

Hence,

$$\int\sqrt{\sin^3x} \;\; dx = \frac{1}{4\sqrt{\sin^3 x}} - \cos x \sqrt{\sin x} \;\; + \text{ const.}$$

Or perhaps a little more neatly,

$$\int\sqrt{\sin^3x} \;\; dx = \frac{1- 4\cos x \sin^2 x}{4\sqrt{\sin^3 x}} \;\; + \text{ const.}$$

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Hats off, I'm impressed. I never even thought of putting it in the form:

$$\int sin(x)\sqrt{\sin(x)}\;dx$$

let alone taking it further. Good stuff if I think or see any more I'll put them up. I'm between courses atm, so I'm trying to keep my eye in. Apologies for being more than a little rusty.

I've not seen the substitution using $t=-\sin(x)\Rightarrow\frac{dt}{dx}=\cos(x)[/tex] on my course, can you explain how that works exactly? I assume the derivative with respect to dt/dx is equivalent but how does that figure exactly? Sorry if this is a stupid question I genuinely haven't seen it before except obviously in various questions and was wondering about it? Last edited: Hootenanny Staff Emeritus Science Advisor Gold Member Hats off, I'm impressed. Thanks for the kind words I'm between courses atm, so I'm trying to keep my eye in. Apologies for being more than a little rusty. There's not need to apologise, strange as it may seem it was a nice break from working I've not seen the substitution using [itex]t=-\sin(x)\Rightarrow\frac{dt}{dx}=\cos(x)[/tex] on my course, can you explain how that works exactly? I assume the derivative with respect to dt/dx is equivalent but how does that figure exactly? Sorry if this is a stupid question I genuinely haven't seen it before except obviously in various questions and was wondering about it? It's not stupid in the slightest, but the substitution I used was, $$t = \sin x$$ Which when differentiated gives, $$\frac{dt}{dx} = \cos x$$ However, the substitution you quoted isn't true, $$\frac{d}{dx}\left(-\sin x\right) = -\cos x \neq \cos x$$ Last edited: $$t = \sin x$$ Which when differentiated gives, $$\frac{dt}{dx} = \cos x$$ However, the substitution you quoted isn't true, $$\frac{d}{dx}\left(-\sin x\right) = -\cos x \neq \cos x$$ Well sorry I think that was more an error in language, I see what you meant I was copying what you wrote but didn't put in the correct sign, my bad. I see thanks for clearing that up. I've never seen that substitution used before, at least on my course, it all helps. What I should of said is [itex] t=\sin(x)\Rightarrow\frac{dt}{dx}=\cos(x)$

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Hootenanny
Staff Emeritus
Gold Member
Well sorry I think that was more an error in language, I see what you meant I was copying what you wrote but didn't put in the correct sign, my bad. I see thanks for clearing that up. I've never seen that substitution used before, at least on my course, it all helps.

What I should of said is $t=\sin(x)\Rightarrow\frac{dt}{dx}=\cos(x)$
No problem. With reference to the original integral, I didn't immediately jump to integration by parts followed by a substitution. At first I was playing around with trigonometric identities and power reduction formulae. Then I tried integration by parts and saw that it would simplify with a substitution.

There's certainly a 'knack' with substitutions, you get used to spotting which ones will simplify the integral. In particular, a ratio of [some function of] sine and cosine is usually a good indicator that a substitution of the form $u=\sin\theta$ or $u=\cos\theta$ is going to simplify things. There isn't really a set of 'allowed' substitutions, it's simply whatever one thinks will simplify a particular integral.

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Yeah I think anyone who says there isn't an art to mathematics is an idiot. I think calculus is one of those areas where you can be really creative with solutions. Especially with integrals, where often a hint from the ether, can get you moving in the right direction. Practice though does indeed make perfect. Like any art, having some talent will only get you so far, sweat will get you further.

Hi SD I think I have it in terms of elementary functions (unless I've made a daft mistake somewhere).

$$\int\sqrt{\sin^3x}\;\; dx = \int \sin x\sqrt{\sin x} \;\;dx$$

We integrate by parts with,

$$u = \sqrt{\sin x} \Rightarrow \frac{du}{dx} = \frac{\cos x}{2\sqrt{\sin x}}$$

$$dv = \sin x \Rightarrow v = -\cos x$$

Hence,

$$\int \sqrt{\sin^3 x} dx = -\cos x\sqrt{\sin x} - \frac{1}{2}\int\frac{\cos x}{\sqrt{\sin x}} dx$$

We now make a substitution,

$$t = \sin x \Rightarrow \frac{dt}{dx} = \cos x$$

And the integral becomes,

$$\int \sqrt{\sin^3 x} dx = -\cos x\sqrt{\sin x} - \frac{1}{2}\int\frac{dt}{\sqrt{t}}$$

$$= -\cos x\sqrt{\sin x} - \frac{1}{2}\left(-\frac{t^{-3/2}}{2}\right) + \text{ const.}$$

$$= -\cos x\sqrt{\sin x} + \frac{1}{4}\left(\sin^{-3/2}\right) + \text{ const.}$$

Hence,

$$\int\sqrt{\sin^3x} \;\; dx = \frac{1}{4\sqrt{\sin^3 x}} - \cos x \sqrt{\sin x} \;\; + \text{ const.}$$

Or perhaps a little more neatly,

$$\int\sqrt{\sin^3x} \;\; dx = \frac{1- 4\cos x \sin^2 x}{4\sqrt{\sin^3 x}} \;\; + \text{ const.}$$
I never noticed this someone else pointed it out but I think he may have a point?

$$\int \sqrt{\sin^3 x} dx = -\cos x\sqrt{\sin x} - \frac{1}{2}\int\frac{\cos x}{\sqrt{\sin x}} dx$$
The expression in the second integral is incorrect (you need to multiply it by $-\cos x$).

You also integrated it incorrectly.

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D H
Staff Emeritus
Actually even with the extra brackets calc101 has a problem. That's really odd? It doesn't usually fall down on easy integrals?
That's because this is not an easy integral. It has no solution in the elementary functions. In particular, this is not a solution:
$$\frac{3}{2x^{-\frac{1}{2}}} \cdot e^{\frac{3}{2} x^{\frac{1}{2}}$$
Of course, if you allow the use of non-elementary functions, you can get a solution, and this is exactly what Mathematica does.

That's because this is not an easy integral. It has no solution in the elementary functions. In particular, this is not a solution:

Of course, if you allow the use of non-elementary functions, you can get a solution, and this is exactly what Mathematica does.
I checked that with Mathcad and it's correct. Are you sure? I mean I'm more willing to trust a human being over Mathcad if you can show me why. :)

$\int \frac{3}{2}x^{-\frac{1}{2}}\cdot e^{\frac{3}{2}x^{-2}}\,dx\rightarrow \frac{3}{2x^{-\frac{1}{2}}} \cdot e^{\frac{3}{2} x^{\frac{1}{2}}\text {+C}$

What function is used to integrate this as mathcad gives that answer?

The above^^ should if you multiply by -cos(x) come out with an eliptic function. So I think that is also not soluble except in non-elementary functions.

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Hootenanny
Staff Emeritus
Gold Member
I never noticed this someone else pointed it out but I think he may have a point?
The expression in the second integral is incorrect (you need to multiply it by $-\cos x$).
You're correct, well these are daft mistakes, I feel rather stupid now
You also integrated it incorrectly.
Again, you're correct. The integrated function should of course be $\sqrt{t}$ rather than $t^{-3/2}$.

With the appropriate correction the integration by parts becomes,

$$\int \sqrt{\sin^3 x} dx = -\cos x\sqrt{\sin x} + \frac{1}{2}\int\frac{\cos^2 x}{\sqrt{\sin x}} dx$$

Which has no elementary anti-derivative. My apologies SD, I feel like a bit of an ass now

Well me too, I never noticed the mistake until someone pointed it out, despite having gone through it at least once.

Hootenanny
Staff Emeritus
Gold Member
Well me too, I never noticed the mistake until someone pointed it out, despite having gone through it at least once.
Well my thanks go to whoever pointed it out

Well my thanks go to whoever pointed it out
Funnily enough I can't say as it was another forum and it would be treason against PF. In fact I kind of did but didn't plagiarised your work, actually with the proviso that it wasn't my work, which turns out was lucky.

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Ok I got a new one from a friend yesterday.

$$\int e^{(-ax^2)}\;dx$$

At first I thought essentially a is an arbitrary constant and so you can cover it with c, but obviously that's crazy talk, then I though what, subsitution so e^(u) and then logging the result? But this doesn't seem to give me the answer.

My maths program gives

$$e^{(-ax^2)}\cdot x+c$$

So then I tried splitting e up, no luck? Insoluble. Imaginary numbers, perhaps?

I also thought the trick I learnt in this forum a couple of days ago, with $e^{(-x^2)}$ and polar co-ordinates would work but that just leaves something like pi with a constant?

An integrator program gives an error function which looks similar to the solution to $e^{(-x^2)}$? :/

Anyone seen this one before?

I am informed by him at least that it is soluble in elementary functions, but it's beyond me, and my shaky rusty old calculus? Any ideas why it equals the above answer and why it's soluble in elementary functions?

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Gib Z
Homework Helper
In short, It isn't. Your friend lied :P

The little trick you learned only applies for the definite integral from 0 to infinity. And yes, we can give the answer in terms of the error function, but if you look up its definition you'll see why thats obvious.

In short, It isn't. Your friend lied :P

The little trick you learned only applies for the definite integral from 0 to infinity. And yes, we can give the answer in terms of the error function, but if you look up its definition you'll see why thats obvious.
Yeah indeed that's what I mean by only in elementary functions, and ? The erf is an approximation using a series is it not?

And I would say so too if he didn't have a PhD in physics. I'm loathe to tell him he's a big pants on fire liar. He may know something I don't.

Still as someone who's just passed A' level (16-18yr) Calculus for University level maths, is he not taking the mickey d'ya think.

By the way my maths program doesn't give that it gives the $\pi$ and erf too, I accidently defined x above and it came out with that obviously.

2.
$$\int\sqrt{\sin^3x}\;\; dx = \frac{1- 4\cos x \sin^2 x}{4\sqrt{\sin^3 x}} \;\; + \text{ const.}$$​

Mind you I gave him the question above, minus the answer obviously, and told him it was soluble in elementary functions, and he spent 1-2 hours trying to solve it before he gave up. So maybe this is payback.

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D H
Staff Emeritus
Ok I got a new one from a friend yesterday.

$$\int e^{(-ax^2)}\;dx$$

At first I thought essentially a is an arbitrary constant and so you can cover it with c, but obviously that's crazy talk, then I though what, subsitution so e^(u) and then logging the result? But this doesn't seem to give me the answer.

My maths program gives

$$e^{(-ax^2)}\cdot x+c$$
I don't know what your maths program is doing, but it is not right. The derivative of $e^{(-ax^2)}\cdot x+c$ is $e^{(-ax^2)}(1-2ax)$. The function $f(x)=e^{-ax^2}$ has no antiderivative in the elementary functions. End of story.

Yeah indeed that's what I mean by only in elementary functions, and ? The erf is an approximation using a series is it not?
Yes, but then exp, sine, cosine, square root are elementary functions are all approximated on your computer or calculator as well. Nothing special here regarding special functions like erf.

I don't know what your maths program is doing, but it is not right. The derivative of $e^{(-ax^2)}\cdot x+c$ is $e^{(-ax^2)}(1-2ax)$. The function $f(x)=e^{-ax^2}$ has no antiderivative in the elementary functions. End of story.
Explained that above, that was an error on my part.

Yes, but then exp, sine, cosine, square root are elementary functions are all approximated on your computer or calculator as well. Nothing special here regarding special functions like erf.
Yes of course, however given the email below, I think he knows something I don't, or it's ruse?

Dr.Friend said:
Here is an integral which you might find nasty. e^{-ax^2}dx , where a is a constant - using latex type notation. I found it nasty, but luckily found what I needed in a couple of textbooks, but not in the form that I was originally looking for. Another with similar problems is ( e^{-ax} )/x . If you wish to look for the solutions to these, remember that the answer that I found is not exactly in the form that I was looking for, but it was actually still useful for what I needed it for, at least for the first one, which was related to the total energy over all space of some fluid or gas.
The second one isn't an issue, as I haven't tackled it yet and it looks like something I've seen before.

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Gib Z
Homework Helper
Look up the definition of the error function on google =] This should make it all painfully clear :P and no its not an elementary function.

EDIT: wikipedia i should have said.

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