MHB Can Differentiation Under the Integral Sign be Used to Solve this Integral?

[Amad27]

Hi,

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

It is supposed to be done by "differentiation under the integral sign" rule. So we being by making a general form,

$$\int_0^1 \frac{\ln(x +1)}{x^a + 1} dx = I(a)$$But before I start anything, I am very clueless because I am fairly new to the method. I have done partial derivatives so that is no problem.

From the top of this, is it even possible to use differentiation under the integral sign rule here?

Thanks

 

[Euge]

Hi,

INT[ln(x +1)/(x^2 + 1) dx] 0 --> 1

It is supposed to be done by "differentiation under the integral sign" rule. So we being by making a general form,

INT[ln(x + 1)/(x^a + 1) dx] = I(a) But before I start anything, I am very clueless because I am fairly new to the method. I have done partial derivatives so that is no problem.

From the top of this, is it even possible to use differentiation under the integral sign rule here?

Thanks

Hi Olok,

First, let me say that it's not necessary to use differentiation under the integral sign to solve this problem. I'll explain why latter. The integral-defined function $I(a)$ can be differentiated under the integral sign, but after differentiation, you will be left with a messy integrand which will be difficult to compute. You would have a better chance if you differentiated $\int_0^1 \ln(xa + 1)/(x^2 + 1)\, dx$ with respect to $a$. You would have gotten rid of the logarithm by doing this.

To solve the integral without differentiation under the integral sign, use the substitution $x = \tan\theta$, $ dx = \sec^2\theta\, d\theta$ to get

$\displaystyle \int_ 0^1 \frac{\ln(x + 1)}{x^2 + 1} = \int_ 0^{\pi/4} \ln(\tan\theta + 1)\, d\theta$.

Now

$\displaystyle \ln(\tan\theta + 1) = \ln\left(\frac{\sin\theta + \cos\theta}{\cos\theta}\right)$

$\displaystyle = \ln\left(\frac{\sqrt{2}\cos\left(\frac{\pi}{4} - \theta\right)}{\cos(\theta)}\right)$

$\displaystyle = \frac{1}{2}\ln 2 + \ln\cos\left(\frac{\pi}{4} - \theta\right) - \ln\cos\theta$.

Thus

$\displaystyle (*) \int_ 0^{\pi/4} \ln(\tan\theta + 1)\, d\theta = \frac{\pi}{8}\ln{2} + \int_ 0^{\pi/4} \ln\cos\left(\frac{\pi}{4} - \theta\right)\, d\theta - \int_ 0^{\pi/4} \ln\cos\theta\, d\theta$.

By symmetry, or by use of the u- substitution $u = \pi/4 - \theta$ the two integrals on the right of $(*)$ are equal. So the integral in on the left of $(*)$ is $\frac{\pi}{8}\ln{2}$, which implies

$\displaystyle \int_ 0^1 \frac{\ln(x+1)}{x^2 + 1}\, dx = \frac{\pi}{8}\ln{2}$

 

[Amad27]

Hi Olok,

First, let me say that it's not necessary to use differentiation under the integral sign to solve this problem. I'll explain why latter. The integral-defined function $I(a)$ can be differentiated under the integral sign, but after differentiation, you will be left with a messy integrand which will be difficult to compute. You would have a better chance if you differentiated $\int_0^1 \ln(xa + 1)/(x^2 + 1)\, dx$ with respect to $a$. You would have gotten rid of the logarithm by doing this.

To solve the integral without differentiation under the integral sign, use the substitution $x = \tan\theta$, $ dx = \sec^2\theta\, d\theta$ to get

$\displaystyle \int_ 0^1 \frac{\ln(x + 1)}{x^2 + 1} = \int_ 0^{\pi/4} \ln(\tan\theta + 1)\, d\theta$.

Now

$\displaystyle \ln(\tan\theta + 1) = \ln\left(\frac{\sin\theta + \cos\theta}{\cos\theta}\right)$

$\displaystyle = \ln\left(\frac{\sqrt{2}\cos\left(\frac{\pi}{4} - \theta\right)}{\cos(\theta)}\right)$

$\displaystyle = \frac{1}{2}\ln 2 + \ln\cos\left(\frac{\pi}{4} - \theta\right) - \ln\cos\theta$.

Thus

$\displaystyle (*) \int_ 0^{\pi/4} \ln(\tan\theta + 1)\, d\theta = \frac{\pi}{8}\ln{2} + \int_ 0^{\pi/4} \ln\cos\left(\frac{\pi}{4} - \theta\right)\, d\theta - \int_ 0^{\pi/4} \ln\cos\theta\, d\theta$.

By symmetry, or by use of the u- substitution $u = \pi/4 - \theta$ the two integrals on the right of $(*)$ are equal. So the integral in on the left of $(*)$ is $\frac{\pi}{8}\ln{2}$, which implies

$\displaystyle \int_ 0^1 \frac{\ln(x+1)}{x^2 + 1}\, dx = \frac{\pi}{8}\ln{2}$

Hi there,

I have done it this way with success, but I was only wondering how it is possible with differentiation under the integral sign. Shall we try it?

Your idea of

$$\int \frac{\ln(ax + 1)}{x^2 + 1}dx$$ is a good idea, WolframAlpha computes this to be,

$$\frac {x}{(x^2+1)(ax+1) }$$

So we proceed with

$$I'(a) = \int_0^1 \frac {x}{(x^2+1)(ax+1)} dx$$

So we apply the FTC to the RHS, but it seems impossible to find the antiderivative...

Any ideas? Thanks

 

[I like Serena]

Hi there,

I have done it this way with success, but I was only wondering how it is possible with differentiation under the integral sign. Shall we try it?

Your idea of

INT[ ln(ax + 1)/x^2 + 1] is a good idea, WolframAlpha computes this to be,

x/(x^2+1)(ax+1)

So we proceed with

I'(a) = INT[x/(x^2+1)(ax+1) dx] 0 --> 1

So we apply the FTC to the RHS, but it seems impossible to find the antiderivative...

Any ideas? Thanks

From:
$$I'(a) = \int_0^1 \frac{x}{(x^2+1)(ax+1)} dx$$
The next step is partial fraction decomposition.
That is, solve
$$\frac{x}{(x^2+1)(ax+1)} = \frac{Ax+B}{x^2+1} + \frac{C}{ax+1}$$
for $A,B$, and $C$.

 

[Amad27]

From:
$$I'(a) = \int_0^1 \frac{x}{(x^2+1)(ax+1)} dx$$
The next step is partial fraction decomposition.
That is, solve
$$\frac{x}{(x^2+1)(ax+1)} = \frac{Ax+B}{x^2+1} + \frac{C}{ax+1}$$
for $A,B$, and $C$.

Hi,

I tried that, it seems VERY difficult though, since be have the lower case "a" involved as well. But here it is,

$$ (Dx + B)(ax + 1) + C(x^2 + 1) = x$$

To avoid confusion let $$A = D$$ from your $$Ax + B$$
$$ (Dx + B)(ax + 1) + Cx^2 + C = x$$

$$ Dax^2 + Dx + Bax + B + Cx^2 + C = x $$

$$ (Da + C)x^2 + (D + Ba)x +(B + C) = x $$

We don't have any $$x^2$$ terms on the RHS, so $$(Da + C) = 0 $$

Facts Here:
$$Da + C = 0$$ \implies $$C = -Da $$
$$D + Ba = 1$$ \implies $$D = 1 - Ba $$
$$B + C = 0 $$ \implies $$ C = -B $$

We can't add anything here because it won't cancel out anything. This is where I get stuck when trying partial fraction decompositions. But here is the rest of my trial...

Add \implies Equation 1 and \implies Equation 3 together to get

$$ 2C = - (Da + B) $$

Adding \implies Equation 2 and \implies Equation 3 we get,

$$ D + C = 1 -B(a + 1) \implies C = -B(a + 1) - D $$

Subbing this, we get,

$$2[-B(a+1) - D] = -(Da + B)] $$

It is still no use, we are still left with three variables.EDITEDITEDIT

So I thought of an idea here,

$$ (ax + 1)(Dx + B) + C(x^2 + 1) = x $$

Let $$ x = 0$$

(1)(B) + C(1) = x

$$B + C = 0 $$

$$B = -C $$

$$ (ax + 1)(Dx - C) + C(x^2 + 1) = x $$

But that does not help either... I don't there is a partial fraction decomposition.

Let x = 1 [we can try]

$$ (a + 1)(D - C) + 2C = 1 $$

$$ aD - aC + D + C = 1 $$

Nothing still.

 

[I like Serena]

You have correctly deduced the system:
$$\begin{cases}
Da+C&=0 \\
D+Ba&=1 \\
B+C&=0
\end{cases}$$

The straight forward method to solve such a system, is to eliminate the variables one by one.
Let me show you.

With your choice
$$C=-Da \tag{1}$$
the system reduces to:
$$\begin{cases}
D+Ba&=1 \\
B-Da&=0
\end{cases}$$

Then, with the choice
$$B=Da \tag{2}$$
the system is further reduced to:
$$\begin{cases}
D+(Da)a&=1 \\
\end{cases}$$
This yields:
$$D = \frac{1}{1+a^2} \tag{3}$$

Now you can substitute $(3)$ back in $(2)$, yielding $B$.
And then substitute $(3)$ back in $(1)$, yielding $C$.

 

[alyafey22]

This http://mathhelpboards.com/calculus-10/generalized-fractional-logarithm-integral-5467.htmlmight be helpful ...

 

[Euge]

Hi Olok,

I see that you're in working with the other method I mentioned in passing. So I'll explain in detail how to get the answer using it. Let

$\displaystyle J(a) = \int_0^1 \frac{\ln(ax + 1)}{x^2 + 1}\, dx$.

Then

$\displaystyle J'(a) = \int_0^1 \frac{x}{(ax + 1)(x^2 + 1)}\, dx$

$\displaystyle = \int_0^a \frac{u}{(u + 1)(u^2 + a^2)}\, du\quad (\text{using u = ax})$

$\displaystyle = \int_0^a \left(1 - \frac{1}{u + 1}\right)\frac{du}{u^2 + a^2}$

$\displaystyle = \int_0^a \frac{du}{u^2 + a^2} - \int_0^a \frac{du}{(u + 1)(u^2 + a^2)}$

$\displaystyle = \int_0^a \frac{du}{u^2 + a^2} - \frac{1}{a^2 + 1}\int_0^a \left(\frac{1}{u + 1} - \frac{u - 1}{u^2 + a^2}\right)\, du$

$\displaystyle = \left(1 - \frac{1}{a^2 + 1}\right)\int_0^a \frac{ du}{u^2 + a^2} - \frac{1}{a^2 + 1}\int_0^a \frac{du}{u + 1} + \frac{1}{a^2 + 1} \int_0^a \frac{u\, du}{u^2 + a^2}$

$\displaystyle = \frac{a}{a^2 + 1}\frac{\pi}{4} - \frac{\ln(a + 1)}{a^2 + 1} + \frac{1}{a^2 + 1}\frac{\ln(2)}{2}$.

Since $J(0) = 0$, the FTC gives

$\displaystyle J(a) = \int_0^a J'(u)\, du = \frac{\pi}{8}\ln(a^2 + 1) - \int_0^a \frac{\ln(u + 1)}{u^2 + 1}\, du + \frac{\ln(2)}{2}\tan^{-1}(a)$.

Setting $a = 1$ and simplifying results in

$\displaystyle J(1) = \frac{\pi}{4}\ln(2) - J(1)$.

Thus $2J(1) = \frac{\pi}{4}\ln(2)$, or $J(1) = \frac{\pi}{8}\ln(2)$.

 

[Amad27]

Hi Olok,

I see that you're in working with the other method I mentioned in passing. So I'll explain in detail how to get the answer using it. Let

$\displaystyle J(a) = \int_0^1 \frac{\ln(ax + 1)}{x^2 + 1}\, dx$.

Then

$\displaystyle J'(a) = \int_0^1 \frac{x}{(ax + 1)(x^2 + 1)}\, dx$

$\displaystyle = \int_0^a \frac{u}{(u + 1)(u^2 + a^2)}\, du\quad (\text{using u = ax})$

$\displaystyle = \int_0^a \left(1 - \frac{1}{u + 1}\right)\frac{du}{u^2 + a^2}$

$\displaystyle = \int_0^a \frac{du}{u^2 + a^2} - \int_0^a \frac{du}{(u + 1)(u^2 + a^2)}$

$\displaystyle = \int_0^a \frac{du}{u^2 + a^2} - \frac{1}{a^2 + 1}\int_0^a \left(\frac{1}{u + 1} - \frac{u - 1}{u^2 + a^2}\right)\, du$

$\displaystyle = \left(1 - \frac{1}{a^2 + 1}\right)\int_0^a \frac{ du}{u^2 + a^2} - \frac{1}{a^2 + 1}\int_0^a \frac{du}{u + 1} + \frac{1}{a^2 + 1} \int_0^a \frac{u\, du}{u^2 + a^2}$

$\displaystyle = \frac{a}{a^2 + 1}\frac{\pi}{4} - \frac{\ln(a + 1)}{a^2 + 1} + \frac{1}{a^2 + 1}\frac{\ln(2)}{2}$.

Since $J(0) = 0$, the FTC gives

$\displaystyle J(a) = \int_0^a J'(u)\, du = \frac{\pi}{8}\ln(a^2 + 1) - J(1) + \frac{\ln(2)}{2}\tan^{-1}(a)$.

Setting $a = 1$ and simplifying results in

$\displaystyle J(1) = \frac{\pi}{4}\ln(2) - J(1)$.

Thus $2J(1) = \frac{\pi}{4}\ln(2)$, or $J(1) = \frac{\pi}{8}\ln(2)$.

Hello,

This approach seems very difficult to understand. I want to keep pursuing this approach as well, I have one question. How'd you get,

$\displaystyle = \int_0^a \frac{du}{u^2 + a^2} - \frac{1}{a^2 + 1}\int_0^a \left(\frac{1}{u + 1} - \frac{u - 1}{u^2 + a^2}\right)\, du$

I can't seem to understand you step at that point. Meanwhile... Can you also take a look at the partial fraction approach. Taking a look at

You have correctly deduced the system:
$$\begin{cases}
Da+C&=0 \\
D+Ba&=1 \\
B+C&=0
\end{cases}$$

The straight forward method to solve such a system, is to eliminate the variables one by one.
Let me show you.

With your choice
$$C=-Da \tag{1}$$
the system reduces to:
$$\begin{cases}
D+Ba&=1 \\
B-Da&=0
\end{cases}$$

Then, with the choice
$$B=Da \tag{2}$$
the system is further reduced to:
$$\begin{cases}
D+(Da)a&=1 \\
\end{cases}$$
This yields:
$$D = \frac{1}{1+a^2} \tag{3}$$

Now you can substitute $(3)$ back in $(2)$, yielding $B$.
And then substitute $(3)$ back in $(1)$, yielding $C$.

So we know that,

$$ D = \frac{1}{1+a^2} $$

We know that $$B = Da $$ so,

$$ B = \frac{a}{1+a^2} $$

We also know that $$ C = -Da $$

$$ C = \frac{-a}{1+a^2}$$Therefore, we know now,

$$\frac{x}{(x^2+1)(ax+1)} = \frac{Dx+B}{x^2+1} + \frac{C}{ax+1}$$
for $A,B$, and $C$.

And that is,

$$ \frac{\frac{x}{1+a^2} +\frac{a}{1+a^2} }{x^2 + 1} +\frac{\frac{-a}{1+a^2}}{ax+1} $$

$$ = \frac{x+a}{(1+a^2)(x^2+1)} - \frac{a}{(1+a^2)(ax+1)} $$

So now we know the partial fraction decomposition the next step is,

$$ \int_{0}^{1} \frac{x+a}{(1+a^2)(x^2+1)} \,dx - \int_{0}^{1} \frac{a}{(1+a^2)(ax+1)}\,dx $$

We have to consider that "a" is simply a constant.

First we find the first part (left part) of the integral including $$x+a$$

$$ \int_{0}^{1} \frac{x+a}{(1+a^2)(x^2+1)} \,dx $$

= $$ \int_{0}^{1} \frac{x}{(1+a^2)(x^2+1)} \,dx + \int_{0}^{1} \frac{a}{(1+a^2)(x^2+1)} \,dx $$

= $$ \frac{1}{(1+a^2)}\int_{0}^{1} \frac{x}{(x^2+1)} \,dx + \frac{a}{(1+a^2)}\int_{0}^{1}\frac{1}{x^2+1} \,dx $$

The second part is easy, the first part is a bit difficult. Actually, no.

Lets figure out
$$ \frac{1}{(1+a^2)}\int_{0}^{1} \frac{x}{(x^2+1)} \,dx $$ first

Let $$ u = x^2 $$
$$\frac{du}{2} = x dx $$
$$ LowerLimit \implies u = 0^2 = 0 $$
$$ UpperLimit \implies u = 1^2 = 1 $$

$$ = \frac{1}{2+2a^2}\int_{0}^{1} \frac{1}{u+1} \,du $$

$$ = \frac{1}{2 + 2a^2}\ln\left({x^2+1}\right) + {C}_{1} $$ Just the antiderivative With all values 0 --> 1 plugged in it, you get,

$$ = \frac{1}{2 + 2a^2}\ln\left({2}\right) $$

Now, the other part of the integral, which is a simple inverse trig.

$$ \frac{a}{(1+a^2)}\int_{0}^{1}\frac{1}{x^2+1} \,dx $$

$$ = \frac{a}{1+a^2}\arctan\left({x}\right) + {C}_{2}$$ [Just the antiderivative, with values in you get,]

$$ = \frac{a}{1+a^2}(\frac{\pi}{4}) $$

Adding both parts you get

$$ = \frac{1}{2(1 + a^2)}\ln\left({2}\right) + \frac{a}{1+a^2}(\frac{\pi}{4}) $$

$$ = \frac{1}{2(1 + a^2)}\ln\left({2}\right) + \frac{2a}{2(1+a^2)}(\frac{\pi}{4}) $$

$$ = \frac{1}{2(1 + a^2)}\left(\ln\left({2}\right)+ \frac{2a\pi}{4}\right) $$

$$ Let a = 1$$ as originally,

$$ = \frac{1}{2(2)}\left(\ln\left({2}\right)+ \frac{2\pi}{4}\right) $$

$$ = \frac{1}{(4)}\left(\ln\left({2}\right)+ \frac{2\pi}{4}\right) $$

$$ = \frac{\ln\left({2}\right)}{4} + \frac{\pi}{8} $$

Which is wrong... ??

 

[Euge]

Hi Olok,

I did give a hint that doing the problem via differentiation under the integral sign will be challenging :)

I don't see what you mean when you suggest that the approach I'm using differs from a partial fraction approach. Using partial fractions just by itself is not enough to find $J(a)$, but I already used partial fractions in my answer. Namely,

$\displaystyle \frac{1}{(u + 1)(u^2 + a^2)} = \frac{1}{a^2 + 1}\left(\frac{1}{u + 1} - \frac{u - 1}{u^2 + a^2}\right) $.

You can check to see if this is right by combining the fractions on the right-hand side to get the fraction on the left-hand side. This should also answer your question about the step you're confused about.

The answer you have is incorrect. There are two main errors. One of the errors you made is forgetting to calculate the third integral,

$\displaystyle \int_0^1 \frac{a}{(1 + a^2)(ax +
1)}\, dx$.

The second error is assuming that $J'(1)$ is the answer, rather than $J(1)$.

 

[Opalg]

You have done the partial fractions calculation correctly, and your answer is correct up to the point where you conclude that $$ \int_0^1\frac{x}{(x^2+1)(ax+1)}dx = \int_{0}^{1} \frac{x+a}{(1+a^2)(x^2+1)}dx - \int_{0}^{1} \frac{a}{(1+a^2)(ax+1)}dx .$$ But you need to remember that this integral represents $I'(a)$. It is the derivative with respect to $a$ of the integral that you eventually want to evaluate.

Next, you calculated the first of the two integrals on the right-hand side, splitting it up as $\displaystyle \int_{0}^{1} \frac{x}{(1+a^2)(x^2+1)}dx + \int_{0}^{1} \frac{a}{(1+a^2)(x^2+1)}dx $ and correctly getting the answer $ \dfrac{1}{2(1 + a^2)}\left(\ln\left({2}\right)+ \dfrac{2a\pi}{4}\right) .$ But you completely forgot the second of the two integrals on the right-hand side, which is $$ - \int_{0}^{1} \frac{a}{(1+a^2)(ax+1)}dx = -\frac{a}{1+a^2}\biggl[\frac1a\ln(ax+1)\biggr]_0^1 = - \frac{\ln(a+1)}{a^2+1}.$$

Putting everything together, you now see that $$I'(a) = \dfrac{1}{1 + a^2}\left(\tfrac12\ln2+ \tfrac14a\pi - \ln(a+1)\right).$$

Now remember that this is still only the derivative of the final answer. You have to integrate it with respect to $a$ in order to complete the calculation. There are two things that you have to be careful about here. The first is to get the constant of integration correct (to do that, think about what happens when $a=0$). More seriously, one of the terms that you need to integrate is $\displaystyle \int \frac{\ln(a+1)}{1+a^2}da.$ That is exactly the problem that we started out with! But fortunately it has a minus sign in front of it here, so you can take to over to the other side of the equation, which then becomes $2I(a)$.

Edit. I had not seen that Euge already commented to make those points.

 

[Amad27]

Hi Olok,

I did give a hint that doing the problem via differentiation under the integral sign will be challenging :)

I don't see what you mean when you suggest that the approach I'm using differs from a partial fraction approach. Using partial fractions just by itself is not enough to find $J(a)$, but I already used partial fractions in my answer. Namely,

$\displaystyle \frac{1}{(u + 1)(u^2 + a^2)} = \frac{1}{a^2 + 1}\left(\frac{1}{u + 1} - \frac{u - 1}{u^2 + a^2}\right) $.

You can check to see if this is right by combining the fractions on the right-hand side to get the fraction on the left-hand side. This should also answer your question about the step you're confused about.

The answer you have is incorrect. There are two main errors. One of the errors you made is forgetting to calculate the third integral,

$displaystyle \int_0^1 \frac{a}{(1 + a^2)(ax +
1)}\, dx$.

The second error is assuming that $J'(1)$ is the answer, rather than $J(1)$.

Thank you, I can't believe I forgot the third integral, I shall calculate that now.

$$\int_{0}^{1} \frac{a}{(a^2+1)(ax+1)} \,dx$$

$$\frac{a}{a^2+1}$$ is just a constant, we can take that out.

$$ = \frac{a}{a^2+1}\int_{0}^{1}\frac{1}{ax+1} \,dx $$

$$ Let. u = ax+1 $$
$$du = a dx \implies dx = \frac{du}{a}$$

$$ = \frac{1}{(a^2+1)}\cdot\int_{0}^{1}\frac{1}{u} \,du $$

$$ = \frac{1}{(a^2+1)}\cdot \ln\left({ax+1}\right) + C_3$$

Evaluated from 0 to 1 you get

$$ = \frac{1}{(a^2+1)}\cdot \ln\left({a+1}\right) $$

So the total integral now would be the addition of the first TWO parts and this ONE part to get:

$$ = \frac{1}{2(1 + a^2)}\ln\left({2}\right) + \frac{a}{1+a^2}(\frac{\pi}{4}) + \frac{\ln\left({a+1}\right)}{1+a^2} $$

It is important to keep in mind that this is only,

$$ J'(a) = \frac{1}{2(1 + a^2)}\ln\left({2}\right) + \frac{a}{1+a^2}(\frac{\pi}{4}) + \frac{\ln\left({a+1}\right)}{1+a^2} $$

We need to find J(1); that is the goal. So now we should integrate again.

$$ J(a) = \int \frac{1}{2(1 + a^2)}\ln\left({2}\right) + \frac{a}{1+a^2}(\frac{\pi}{4}) + \frac{\ln\left({a+1}\right)}{1+a^2} da $$

$$ = \frac{\ln\left({2}\right)}{2}\cdot\int\frac{1}{1+a^2} da + \frac{\pi}{4}\cdot\int\frac{a}{1+a^2} da + \int \frac{\ln\left({1+a}\right)}{1+a^2} da $$

$$ = \frac{\ln\left({2}\right)}{2}\cdot\arctan\left({a}\right) + \frac{\pi}{8}\cdot\ln\left({a^2+1}\right) + \int \frac{ln(1+a)}{1+a^2} da $$

The difficult part is $$\int \frac{ln(1+a)}{1+a^2} da $$

That is the issue, that doesn't have an elementary antiderivative. Any ideas?

- - - Updated - - -

Hi Olok,

I did give a hint that doing the problem via differentiation under the integral sign will be challenging :)

I don't see what you mean when you suggest that the approach I'm using differs from a partial fraction approach. Using partial fractions just by itself is not enough to find $J(a)$, but I already used partial fractions in my answer. Namely,

$\displaystyle \frac{1}{(u + 1)(u^2 + a^2)} = \frac{1}{a^2 + 1}\left(\frac{1}{u + 1} - \frac{u - 1}{u^2 + a^2}\right) $.

You can check to see if this is right by combining the fractions on the right-hand side to get the fraction on the left-hand side. This should also answer your question about the step you're confused about.

The answer you have is incorrect. There are two main errors. One of the errors you made is forgetting to calculate the third integral,

$displaystyle \int_0^1 \frac{a}{(1 + a^2)(ax +
1)}\, dx$.

The second error is assuming that $J'(1)$ is the answer, rather than $J(1)$.

Thank you, I can't believe I forgot the third integral, I shall calculate that now.

$$\int_{0}^{1} \frac{a}{(a^2+1)(ax+1)} \,dx$$

$$\frac{a}{a^2+1}$$ is just a constant, we can take that out.

$$ = \frac{a}{a^2+1}\int_{0}^{1}\frac{1}{ax+1} \,dx $$

$$ Let. u = ax+1 $$
$$du = a dx \implies dx = \frac{du}{a}$$

$$ = \frac{1}{(a^2+1)}\cdot\int_{0}^{1}\frac{1}{u} \,du $$

$$ = \frac{1}{(a^2+1)}\cdot \ln\left({ax+1}\right) + C_3$$

Evaluated from 0 to 1 you get

$$ = \frac{1}{(a^2+1)}\cdot \ln\left({a+1}\right) $$

So the total integral now would be the addition of the first TWO parts and this ONE part to get:

$$ = \frac{1}{2(1 + a^2)}\ln\left({2}\right) + \frac{a}{1+a^2}(\frac{\pi}{4}) + \frac{\ln\left({a+1}\right)}{1+a^2} $$

It is important to keep in mind that this is only,

$$ J'(a) = \frac{1}{2(1 + a^2)}\ln\left({2}\right) + \frac{a}{1+a^2}(\frac{\pi}{4}) + \frac{\ln\left({a+1}\right)}{1+a^2} $$

We need to find J(1); that is the goal. So now we should integrate again.

$$ J(a) = \int \frac{1}{2(1 + a^2)}\ln\left({2}\right) + \frac{a}{1+a^2}(\frac{\pi}{4}) + \frac{\ln\left({a+1}\right)}{1+a^2} da $$

$$ = \frac{\ln\left({2}\right)}{2}\cdot\int\frac{1}{1+a^2} da + \frac{\pi}{4}\cdot\int\frac{a}{1+a^2} da + \int \frac{\ln\left({1+a}\right)}{1+a^2} da $$

$$ = \frac{\ln\left({2}\right)}{2}\cdot\arctan\left({a}\right) + \frac{\pi}{8}\cdot\ln\left({a^2+1}\right) + \int \frac{ln(1+a)}{1+a^2} da $$

The difficult part is $$\int \frac{ln(1+a)}{1+a^2} da $$

That is the issue, that doesn't have an elementary antiderivative. Any ideas?

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Hi Olok,

I did give a hint that doing the problem via differentiation under the integral sign will be challenging :)

I don't see what you mean when you suggest that the approach I'm using differs from a partial fraction approach. Using partial fractions just by itself is not enough to find $J(a)$, but I already used partial fractions in my answer. Namely,

$\displaystyle \frac{1}{(u + 1)(u^2 + a^2)} = \frac{1}{a^2 + 1}\left(\frac{1}{u + 1} - \frac{u - 1}{u^2 + a^2}\right) $.

You can check to see if this is right by combining the fractions on the right-hand side to get the fraction on the left-hand side. This should also answer your question about the step you're confused about.

The answer you have is incorrect. There are two main errors. One of the errors you made is forgetting to calculate the third integral,

$displaystyle \int_0^1 \frac{a}{(1 + a^2)(ax +
1)}\, dx$.

The second error is assuming that $J'(1)$ is the answer, rather than $J(1)$.

Thank you, I can't believe I forgot the third integral, I shall calculate that now.

$$\int_{0}^{1} \frac{a}{(a^2+1)(ax+1)} \,dx$$

$$\frac{a}{a^2+1}$$ is just a constant, we can take that out.

$$ = \frac{a}{a^2+1}\int_{0}^{1}\frac{1}{ax+1} \,dx $$

$$ Let. u = ax+1 $$
$$du = a dx \implies dx = \frac{du}{a}$$

$$ = \frac{1}{(a^2+1)}\cdot\int_{0}^{1}\frac{1}{u} \,du $$

$$ = \frac{1}{(a^2+1)}\cdot \ln\left({ax+1}\right) + C_3$$

Evaluated from 0 to 1 you get

$$ = \frac{1}{(a^2+1)}\cdot \ln\left({a+1}\right) $$

So the total integral now would be the addition of the first TWO parts and this ONE part to get:

$$ = \frac{1}{2(1 + a^2)}\ln\left({2}\right) + \frac{a}{1+a^2}(\frac{\pi}{4}) + \frac{\ln\left({a+1}\right)}{1+a^2} $$

It is important to keep in mind that this is only,

$$ J'(a) = \frac{1}{2(1 + a^2)}\ln\left({2}\right) + \frac{a}{1+a^2}(\frac{\pi}{4}) + \frac{\ln\left({a+1}\right)}{1+a^2} $$

We need to find J(1); that is the goal. So now we should integrate again.

$$ J(a) = \int \frac{1}{2(1 + a^2)}\ln\left({2}\right) + \frac{a}{1+a^2}(\frac{\pi}{4}) + \frac{\ln\left({a+1}\right)}{1+a^2} da $$

$$ = \frac{\ln\left({2}\right)}{2}\cdot\int\frac{1}{1+a^2} da + \frac{\pi}{4}\cdot\int\frac{a}{1+a^2} da + \int \frac{\ln\left({1+a}\right)}{1+a^2} da $$

$$ = \frac{\ln\left({2}\right)}{2}\cdot\arctan\left({a}\right) + \frac{\pi}{8}\cdot\ln\left({a^2+1}\right) + \int \frac{ln(1+a)}{1+a^2} da $$

The difficult part is $$\int \frac{ln(1+a)}{1+a^2} da $$

That is the issue, that doesn't have an elementary antiderivative. Any ideas?

 

[Amad27]

You have done the partial fractions calculation correctly, and your answer is correct up to the point where you conclude that $$ \int_0^1\frac{x}{(x^2+1)(ax+1)}dx = \int_{0}^{1} \frac{x+a}{(1+a^2)(x^2+1)}dx - \int_{0}^{1} \frac{a}{(1+a^2)(ax+1)}dx .$$ But you need to remember that this integral represents $I'(a)$. It is the derivative with respect to $a$ of the integral that you eventually want to evaluate.

Next, you calculated the first of the two integrals on the right-hand side, splitting it up as $\displaystyle \int_{0}^{1} \frac{x}{(1+a^2)(x^2+1)}dx + \int_{0}^{1} \frac{a}{(1+a^2)(x^2+1)}dx $ and correctly getting the answer $ \dfrac{1}{2(1 + a^2)}\left(\ln\left({2}\right)+ \dfrac{2a\pi}{4}\right) .$ But you completely forgot the second of the two integrals on the right-hand side, which is $$ - \int_{0}^{1} \frac{a}{(1+a^2)(ax+1)}dx = -\frac{a}{1+a^2}\biggl[\frac1a\ln(ax+1)\biggr]_0^1 = - \frac{\ln(a+1)}{a^2+1}.$$

Putting everything together, you now see that $$I'(a) = \dfrac{1}{1 + a^2}\left(\tfrac12\ln2+ \tfrac14a\pi - \ln(a+1)\right).$$

Now remember that this is still only the derivative of the final answer. You have to integrate it with respect to $a$ in order to complete the calculation. There are two things that you have to be careful about here. The first is to get the constant of integration correct (to do that, think about what happens when $a=0$). More seriously, one of the terms that you need to integrate is $\displaystyle \int \frac{\ln(a+1)}{1+a^2}da.$ That is exactly the problem that we started out with! But fortunately it has a minus sign in front of it here, so you can take to over to the other side of the equation, which then becomes $2I(a)$.

Edit. I had not seen that Euge already commented to make those points.

Excellent, the sign fixed my mistake, so I have this now.

There is still a problem, remember that,

$$I(a) = \int_{0}^{1} \frac{\ln\left({ax+1}\right)}{x^2+1} \,dx$$

What you pointed out here,

$\displaystyle \int \frac{\ln(a+1)}{1+a^2}da.$

Is not of much use. If we add the first and the second, we do not get $$2I(a)$$

Moreover, what you point out,

$\displaystyle \int \frac{\ln(a+1)}{1+a^2}da \ne I(a)$

 

[Opalg]

There is still a problem, remember that,

$$I(a) = \int_{0}^{1} \frac{\ln\left({ax+1}\right)}{x^2+1} \,dx$$

What you pointed out here,

$\displaystyle \int \frac{\ln(a+1)}{1+a^2}da.$

Is not of much use. If we add the first and the second, we do not get $$2I(a)$$

Moreover, what you point out,

$\displaystyle \int \frac{\ln(a+1)}{1+a^2}da \ne I(a)$

That was me being careless. What you end up with here is not twice $I(a)$ but twice $I(1)$, because to get the final result you need to put $a=1$.

 

[Euge]

Hello again Olok,

In your formula for $J(a)$, the last tern should have a minus sign out front, rather than a plus sign. Before you integrate $J'(a)$, find an initial value. The easiest value to find is $J(0) = 0$. Now you can use FTC to find $J(1)$:

$\displaystyle J(1) = \int_0^1 J'(a)\, da$.

What is crucial here is that

$J(1) = \int_0^1 \frac{\ln(a + 1)}{a^2 + 1}\, da$.

This is why the answer to the original problem is the value of $J(1)$. After integration you should get the equation

$\displaystyle J(1) = \frac{\pi}{4}\ln(2) - J(1)$,

which implies $J(1) = \frac{\pi}{8}\ln(2)$.

 

[Amad27]

That was me being careless. What you end up with here is not twice $I(a)$ but twice $I(1)$, because to get the final result you need to put $a=1$.

Ok, I need a moment to soak this in...

$$I(a)$$ as we defined it is,

$$I(a) = \int_{0}^{1} \frac{\ln\left({ax+1}\right)}{x^2+1}\,dx$$

So

$$I(1) = \int_{0}^{1} \frac{\ln\left({x+1}\right)}{x^2+1} \,dx $$

$$\int_{0}^{1} \frac{\ln\left({x+1}\right)}{x^2+1} \,dx \ne \int_{0}^{1}\frac{ln(a+1)}{1+a^2} \,da $$

Because here, $$ a \ne x$$

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...

What is crucial here is that

$J(1) = \int_0^1 \frac{\ln(a + 1)}{a^2 + 1}\, da$.
...

But HOW is J(1) that? As you define J(a),

$$J(a) = \int_{0}^{1} \frac{\ln\left({ax+1}\right)}{x^2+1} \,dx $$

$$J(1) = \int_{0}^{1} \frac{\ln\left({x+1}\right)}{x^2+1} \,dx $$

As you type here you mean to say that,

$J(1) = \int_0^1 \frac{\ln(a + 1)}{a^2 + 1}\, da$.

But it is important to realize that,

$J(1) = \int_0^1 \frac{\ln(a + 1)}{a^2 + 1}\, da \ne \int_{0}^{1} \frac{\ln\left({x+1}\right)}{x^2+1} \,dx $

 

[Euge]

I think you already know this, but in the formula

$\displaystyle J(a) = \int_0^1 \frac{\ln(ax + 1)}{x^2 + 1}\, dx$,

the $a$ is not a dummy variable. However, in the integral

$\displaystyle \int_0^1 \frac{\ln(a + 1)}{a^2 + 1}\, da$,

the $a$ is a dummy variable.

 

[Amad27]

I think you already know this, but in the formula

$\displaystyle J(a) = \int_0^1 \frac{\ln(ax + 1)}{x^2 + 1}\, dx$,

the $a$ is not a dummy variable. However, in the integral

$\displaystyle \int_0^1 \frac{\ln(a + 1)}{a^2 + 1}\, da$,

the $a$ is a dummy variable.

Just a question,

BEFORE, a is not the dummy variable,
AFTER, a is the dummy variable.

Something doesn't seem right here...

 

[Euge]

If it helps you to understand, after you find J'(a), change all the $a$'s to an $x$, then integrate.

 

[Amad27]

If it helps you to understand, after you find J'(a), change all the $a$'s to an $x$, then integrate.

Hi,

It makes good sense to just call it a dummy variable because it is the

$$da$$ which tells you that it is a dummy variable just like when you

Have $$dt$$ and you replace the variables with $$dx$$

I have another interesting integral, which also is a differentiation under the integral sign, which I am posting soon.