Cracking the Tough Integral: \int_0^1 \frac{ln(x+1)}{x^2+1}?

[Jungy]

Homework Statement

\int_0^1 \frac{ln(x+1)}{x^2+1}

The Attempt at a Solution

Alright so I've come across this integral as the last question in the whole exercise set, and... I've gotten nothing...
All previous questions have involved some sort of substitution.. but I can't find any suitable subs for this one.
By parts hasn't gotten anything for me; you get a tan inverse and that becomes another wall.
An answer has been provided; \frac{1}{8} \pi ln2... but I have no idea how you get there.

 

[snipez90]

Try substituting x = tan(t) and further manipulate the numerator.

 

[Jungy]

Let's go...
x = tan\,u
dx = sec^2\,u\,du

so it becomes
\int_0^\frac{\pi}{4} ln[tan(u)+1]\,du

and now I'm stuck again...

 

[kof9595995]

\int_0^\frac{\pi}{4} ln[tan(u)+1]\,du
Since you've got here, try substitute t=\frac{\pi}{4}-u

 

[Jungy]

After a bit of fiddling:

\int_0^\frac{\pi}{4} ln[tan(\frac{\pi}{4}-t)+1)\,dt

I'm still not picking up anything... by parts from here doesn't look to friendly

 

[Dick]

If it's any help, and it's not, I don't see it either. I usually can and I don't. I'm just posting to get an alert if someone knows the secret trick. Don't feel bad for not getting it.

 

[snipez90]

Yes, well that's the general idea: don't use integration by parts. The first substitution was natural, and simplified the integral quite a bit. As for the resulting integral, the log function alone suggests that we are not going to do the integral by parts. But more importantly, the nice limits of integration and the appearance of the trig function suggest that we exploit the symmetry of trig functions. Thus, at this point, you should realize that clearly, there is justification in kof9595995's suggested substitution. In fact, try to prove why that \int_{a}^{b}f(a + b - x) \,dx = \int_{a}^{b}f(x) \,dx. This is a useful trick.

Expand tan(a-b), and that should lead you to the solution. Also, when doing definite integration, it's almost always a good idea to denote the integral trying to determine by I or some other letter from the beginning. You could also do this integral without using kof9595995's substitution. The more straightforward way is to simply write tan(x) + 1 as [sin(x) + cos(x)] / cos(x). The expression sin(x) + cos(x) can be rewritten as A*sin(x + B), for suitable A and B. A few more substitutions will lead to the answer.

 

[Jungy]

Thanks for all the replies + Dick's input (lol)

Okay so I'm just going to spell out everything I've tried doing so far:

Using kof's substitution ->
Expanding tan\,(\frac{\pi}{4}-t)
I get \frac{1-tan\,t}{1+tan\,t}
The integral of that, I recognised is ln (cos+sin) if that's of any use later on..?

Not sure how to deal with this if there's a natural log in front of it to integrate... and I don't see it as very useful, as nice as it looks... or maybe I'm just braindead and forgetting something really obvious?

Going back to snipez90's suggestion of tan u + 1 = that sin, cos combo;
i get down to \int_0^\frac{\pi}{4} ln\,(sin\,u+cos\,u)-ln\,(cos\,u)\,du

I sort of understand the "suitable A, B to get A*sin(x+B)," for example,
\sqrt{2}sin(u+\frac{\pi}{4}) gives sin u + cos u, although I still don't get how to incorporate this because of that natural log... argh!

 

[FunkyDwarf]

This is probably rather unhelpful (only had a quick glance so far) but if you complete the square on the bottom you get an x+1 squared term (and a -2x term which is bad) but perhaps that would help in the by parts method? I don't know =(

 

[snipez90]

For the first approach, you are forgetting something obvious, namely a 1. Then see if you can write the original integral (give it a variable to keep track of it if you haven't) as a difference of two integrals (like you did in the second approach).

Well for the second approach, the log actually helps, since sqrt(2) is just a constant after all. Note that you actually get the answer you want once you split off the ln(sqrt(2)) by the linearity property of integrals. This should tell you how to finish the solution, since you have 2 other integrals that are "left over".

 

[nickmai123]

Haha I love the result Maple gave me. What's a dilog?

EDIT: Nvm the dilog. One of those non-elementary functions.

 

[Jungy]

Ah yes, snipez, silly me, wrote it down and forgot to post it up here
it becomes 2/(1+tan (t));

so we get integral ln 2 - ln (1+tan t) but i still don't like the integral ln (1+tan t) bit... how would I go about doing that...

with the second approach:
If you're saying what I think you're saying:
integral ln(sqrt2)+ ln(sin(u+pi/4) -ln(cos u) [or + ln(sin(u-pi/2)??] du

Again I'm stuck with doing integral natural logs of trig..
Or am I just going in the wrong direction?

 

[kof9595995]

Since you've got
<br /> \int_0^\frac{\pi}{4} ln[tan(u)+1]\,du=\int_0^\frac{\pi}{4} ln2-ln[tan(t)+1]\,dt<br />
Isn't it obvious that
<br /> 2\int_0^\frac{\pi}{4} ln[tan(u)+1]\,du=\int_0^\frac{\pi}{4} ln2\,dt<br />
Then you know what to do next.

 

[Jungy]

dummy variables, solves having to integrate log trigs ^^;
And it becomes so simple after all that work, with an easy integration to end.
Hahaha I'm so slow...

Thanks kof + everyone else who contributed

 

[snipez90]

Yeah, that is why it is a good idea to label your integrals throughout.

Jungy, your algebraic manipulations are fine, but I think you are still having trouble with the concepts (which is fine, since this is a hard integral). The solutions outlined here do not attempt to actually evaluate any integrals of log functions composed with trig functions directly. Instead, the trick is to manipulate the limits of integration to break down the original integral and simplify. The best thing about definite integrals involving trig functions is that we can exploit symmetry, often by making simple substitutions. For instance, if we are integrating an expression involving a trig function with respect to x over the interval [0, pi/4], we might simply try to substitute u = pi/4 - x to see if the "new" integral preserves any nice properties of the original integral. Moreover, various trig identities, such as sin(pi/2 - x) = cos(x) for real x may serve as suggestions for which substitutions we make.

In the end, definite integrals of this type involve some general principles (such as the integral identity I mentioned above), but keeping the basic idea in mind is more important. Having said that, you really shouldn't let integrals of logarithms of trig functions intimidate you. You can actually draw out the graphs of sin(pi/4 + x) and cos(x) and see that taking their integrals over the interval 0 to pi/4 give you the same answer. You should prove this by making the appropriate substitution (or noticing that you could write sin(x) + cos(x) can be rewritten in terms of cos as well).

But really, if you have a definite integral and obvious substitutions don't work and integrating by parts looks ugly, it's pretty much given that the limits of integration play a huge role. Ordinarily, all you have to do is remember to change the limits of integration. You can actually evaluate this integral using direct methods, but these approaches are more advanced. Differentiating under the integral sign works. Applying series expansion works (ugly). Contour integration probably works, but someone who knows complex analysis will help you out there.

 

[Jungy]

Yeah I have 'I=' in front of my work all the time, just didn't put it here... although you rightly pose an important point..
I'm not 100% confident about freely fiddling with the limits, although I know those 'properties of definite integrals' or whatnot, and are obviously relevant, especially in cases like this..
I have definitely learned from this... thanks again to you all

Now to the next chapter: resisted motion :D

 

[vm310]

\int_0^\frac{\pi}{4} ln[tan(u)+1]\,du
Since you've got here, try substitute t=\frac{\pi}{4}-u

Sorry to bring up this old post, but I was wondering how you decided to use this substitution?

Thanks

 

[Gib Z]

Either by having seen the beautiful solution before and having remembered it, or by trying his luck with the integral identity \int^a_0 f(x) dx = \int^a_0 f(a-x) dx because the lower bound was zero, or just perhaps having the foresight to realize the nice manipulation that can come about from the substitution.

 

[vm310]

\int^a_0 f(x) dx = \int^a_0 f(a-x) dx

I'm sorry but I can't find this integral identity anywhere, not in my book or online. Is this the standard form for the identity or has this a manipulation of one?

Thanks so much

 

[Gib Z]

That is the standard form of that identity. You can show it by letting x= a-u in the Left Integral and by recognising that x and u are just dummy variables in the definite integral and change back to x at the end.