How to Evaluate This Integral Using a Table of Integrals?

[EngWiPy]

Hello all,

I need to evaluate this integral

\int_0^{\infty}(1+\alpha_2)^{-2}\left(\alpha_2+b\right)^{-1}\,d\alpha_2

I found the following integral in the table of integrals (eq 3.197.1, 7th edition)

\int_0^{\infty}x^{v-1}(\beta+x)^{-\mu}(x+\gamma)^{-\rho}\,dx=\beta^{-\mu}\gamma^{\mu-\rho}B(v,\mu-v+\rho)_2F_1(\mu,\,v;\,\mu+\rho;1-\frac{\gamma}{\beta})

The integral and the conditions are in the image attached. Comparing the two integrals I set $v=1,\,\beta=1,\,\gamma=b$, which implies that $\mu=2$ and $\rho=1$. I satisfy the last two conditions, but I'm not sure about the first two. What does it mean arg x if x is a constant?

$b$ varies in a loop, and when I evaluate the integral I get results. However, the final result, in which the above integral is a part, isn't right. I'm not sure where the error is, and I wanted to make sure it isn't in using the integral.

I appreciate any help. Thanks

 

[Mark44]

Hello all,

I need to evaluate this integral

\int_0^{\infty}(1+\alpha_2)^{-2}\left(\alpha_2+b\right)^{-1}\,d\alpha_2

I found the following integral in the table of integrals (eq 3.197.1, 7th edition)

\int_0^{\infty}x^{v-1}(\beta+x)^{-\mu}(x+\gamma)^{-\rho}\,dx=\beta^{-\mu}\gamma^{\mu-\rho}B(v,\mu-v+\rho)_2F_1(\mu,\,v;\,\mu+\rho;1-\frac{\gamma}{\beta})

The integral and the conditions are in the image attached. Comparing the two integrals I set $v=1,\,\beta=1,\,\gamma=b$, which implies that $\mu=2$ and $\rho=1$. I satisfy the last two conditions, but I'm not sure about the first two. What does it mean arg x if x is a constant?

It doesn't say arg(x) in the integral you found -- it says arg(β) and arg(γ). This implies that these parameters are complex. In the integral x is real.

$b$ varies in a loop, and when I evaluate the integral I get results. However, the final result, in which the above integral is a part, isn't right. I'm not sure where the error is, and I wanted to make sure it isn't in using the integral.

I appreciate any help. Thanks

The integral you found should work, but I don't know what the B and F₁ functions are. The integral could be evaluated using much simpler means, though, by using partial fractions decomposition to split the integrand into three parts.

The decomposition would look like this:
$$\frac{1}{(a + 1)^2} \frac{1}{a + b} = \frac{A}{a + 1} + \frac{B}{(a + 1)^2} + \frac{C}{a + b} $$
After you determine the constants A, B, and C, integrate each term and then evaluate your three improper integrals.

 

[mfb]

The equation is valid for some complex numbers as well, the condition on the complex argument are satisfied for positive real values (and also for all complex values with a positive imaginary component).

It is not necessary to use such a powerful formula, for integer exponents you can use partial fraction decomposition.

 

[EngWiPy]

...The integral could be evaluated using much simpler means, though, by using partial fractions decomposition to split the integrand into three parts.

The decomposition would look like this:
$$\frac{1}{(a + 1)^2} \frac{1}{a + b} = \frac{A}{a + 1} + \frac{B}{(a + 1)^2} + \frac{C}{a + b} $$
After you determine the constants A, B, and C, integrate each term and then evaluate your three improper integrals.

Thanks, but why $\frac{A}{a + 1} + \frac{B}{(a + 1)^2} + \frac{C}{a + b}$ and not $\frac{A}{(a + 1)^2} + \frac{B}{a + b}$? I did the latter, and I got the following:

\frac{1}{(x+1)^2(x+b)}=\frac{1}{b-1}\left[\frac{1}{(x+1)^2}-\frac{1}{x+b}\right]

Then I used the integral formula (attached):

\int_0^{\infty}\frac{x^{\mu-1}}{(1+\beta\,x)^v}\,dx=\beta^{-\mu} B(\mu,\,v-\mu),

where $B(.,.)$ is the beta function. I got the following:

\int_0^{\infty}\frac{1}{(x+1)^2}\,dx=B(1,1)

and

\int_0^{\infty}\frac{1}{(x+b)}\,dx=B(1,0)

But evaluating $B(1,1)-B(1,0)$ in Mathematica gives me ComplexInfinity! Why?

 

[Mark44]

Thanks, but why $\frac{A}{a + 1} + \frac{B}{(a + 1)^2} + \frac{C}{a + b}$ and not $\frac{A}{(a + 1)^2} + \frac{B}{a + b}$?

Because that's how you do it when you have repeated linear factors (i.e., (a + 1)²). If you break it up into only two terms, you don't get the correct decomposition. You need to break the original integrand into three terms, not two.

I did the latter, and I got the following:

\frac{1}{(x+1)^2(x+b)}=\frac{1}{b-1}\left[\frac{1}{(x+1)^2}-\frac{1}{x+b}\right]

Then I used the integral formula (attached):

\int_0^{\infty}\frac{x^{\mu-1}}{(1+\beta\,x)^v}\,dx=\beta^{-\mu} B(\mu,\,v-\mu),

where $B(.,.)$ is the beta function. I got the following:

\int_0^{\infty}\frac{1}{(x+1)^2}\,dx=B(1,1)

and

\int_0^{\infty}\frac{1}{(x+b)}\,dx=B(1,0)

But evaluating $B(1,1)-B(1,0)$ in Mathematica gives me ComplexInfinity! Why?

 

[EngWiPy]

Because that's how you do it when you have repeated linear factors (i.e., (a + 1)²). If you break it up into only two terms, you don't get the correct decomposition. You need to break the original integrand into three terms, not two.

Actually, I did it as you said, and got A=0, which led to the same result as I did it!

 

[Mark44]

Actually, I did it as you said, and got A=0, which led to the same result as I did it!

Then you made a mistake -- A is not 0. I'm pretty confident in this, as I checked my work.

If you show what you did, we can figure out where you went wrong.

 

[EngWiPy]

Then you made a mistake -- A is not 0. I'm pretty confident in this, as I checked my work.

If you show what you did, we can figure out where you went wrong.

OK. So first we have

<br /> \frac{1}{(x + 1)^2} \frac{1}{x + b} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{C}{x + b}=\frac{A(x+1)(x+b)+B(x+b)+C(x+1)^2}{(x+1)^2(x+b)}<br />

This implies that

$$A(x+1)(x+b)+B(x+b)+C(x+1)^2=1$$

For $x=-1$ we get $B=\frac{1}{b-1}$

For $x=-b$ we get $C=-\frac{1}{b-1}$

For $x=0$ we get $Ab+Bb+C=Ab+\frac{b}{b-1}-\frac{1}{b-1}=Ab+1=1$ which implies that $A=0$.

Where did I err?

 

[Mark44]

OK. So first we have

<br /> \frac{1}{(x + 1)^2} \frac{1}{x + b} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{C}{x + b}=\frac{A(x+1)(x+b)+B(x+b)+C(x+1)^2}{(x+1)^2(x+b)}<br />

This implies that

$$A(x+1)(x+b)+B(x+b)+C(x+1)^2=1$$

For $x=-1$ we get $B=\frac{1}{b-1}$

For $x=-b$ we get $C=-\frac{1}{b-1}$

For $x=0$ we get $Ab+Bb+C=Ab+\frac{b}{b-1}-\frac{1}{b-1}=Ab+1=1$ which implies that $A=0$.

Where did I err?

In your calculation for C. I get C = $-\frac{1}{(b - 1)^2}$

 

[EngWiPy]

In your calculation for C. I get C = $-\frac{1}{(b - 1)^2}$

I think in this case $C=\frac{1}{(b-1)^2}$, isn't it? But you are right, I missed the square, and don't know how I missed it THREE times! Anyway, thanks

 

[Mark44]

I think in this case $C=\frac{1}{(b-1)^2}$, isn't it?

Correct. I misread my work.

But you are right, I missed the square, and don't know how I missed it THREE times! Anyway, thanks

One little mistake will throw everything off. That's why I checked my work. After substituting the values for A, B, and C, I ended up with $\frac{(b - 1)^2}{(b - 1)^2} \equiv 1$ (except if b = 1).

 

[EngWiPy]

The integral I attached in post #4 works only for $(x+1)^2$. How to evaluate the other integrals? I found this indefinite integral

$$\int\frac{1}{ax+b}\,dx=\frac{1}{a}\ln|ax+b|$$

Can I use this?

 

[Mark44]

The integral I attached in post #4 works only for $(x+1)^2$. How to evaluate the other integrals? I found this indefinite integral

$$\int\frac{1}{ax+b}\,dx=\frac{1}{a}\ln|ax+b|$$

Can I use this?

Yes. For both integrals a = 1.

 

[EngWiPy]

Yes. For both integrals a = 1.

Right, but the integrals in my case are definite. Do I need to evaluate the result as $\left. \frac{1}{a}\ln|ax+b|\right|_0^{\infty}$. This will result in $+\infty$!

 

[Mark44]

All three are improper integrals, so you can't simply substitute $\infty$ into the antiderivatives. You'll need to use limits. Any calculus textbook should have a section on dealing with improper integrals.