Solve Integral w/ Log: $$\int_{5}^{\infty}\frac{1}{t^r\log t}dt$$

In summary, the integral $$\int_{5}^{\infty}\frac{1}{\log\left(t\right)t^{r}}dt$$ with $r\geq 2$ can be solved using the substitution $x = (r-1)\log t$, which transforms it into the exponential integral $\int_{(r-1)\log 5}^\infty \frac{e^{-x}}xdx$. However, this integral cannot be expressed in terms of elementary functions.
  • #1
Bibubo
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How can I solve (or evaluating) this integral $$\int_{5}^{\infty}\frac{1}{\log\left(t\right)t^{r}}dt$$ with $r\geq 2$?
 
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  • #2
Bibubo said:
How can I solve (or evaluating) this integral $$\int_{5}^{\infty}\frac{1}{\log\left(t\right)t^{r}}dt$$ with $r\geq 2$?

I would expect you need to use integration by parts...
 
  • #3
$$\int^\infty_5 \frac{1}{\log(t)\,t^r}\,dt$$

Let $\log(t) = x \implies \,t=e^x \,\,\,\,;dt=e^x \,dx$

$$\int^\infty_{\log(5)} \frac{e^{-x(r-1)}}{x}\,dt$$

Let $z= x(r-1) \,\, \implies dz = r-1\,dz$

$$\int^\infty_{\log(5)(r-1)} \frac{e^{-z}}{z}\,dz = E_1\left(\log(5)(r-1) \right)$$

Where the exponential integral is defined

$$E_n(x) = \int^\infty_1 \frac{e^{-xt}}{t^n}\,dt$$
 
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  • #4
Bibubo said:
How can I solve (or evaluating) this integral $$\int_{5}^{\infty}\frac{1}{\log\left(t\right)t^{r}}dt$$ with $r\geq 2$?
The substitution $x = (r-1)\log t$ transforms this into the exponential integral \(\displaystyle \int_{(r-1)\log 5}^\infty \frac{e^{-x}}xdx\), which cannot be expressed in terms of elementary functions.

Edit. Sorry, didn't see ZaidAlyafey's comment.
 
  • #5
ZaidAlyafey said:
$$\int^\infty_5 \frac{1}{\log(t)\,t^r}\,dt$$

Let $\log(t) = x \implies \,t=e^x \,\,\,\,;dt=e^x \,dx$

$$\int^\infty_{\log(5)} \frac{e^{-x(r-1)}}{x}\,dt$$

Let $t= x(r-1) \,\, \implies dt = r-1\,dx$

$$\int^\infty_{\log(5)(r-1)} \frac{e^{-t}}{t}\,dt = E_1\left(\log(5)(r-1) \right)$$

Where the exponential integral is defined

$$E_n(x) = \int^\infty_1 \frac{e^{-xt}}{t^n}\,dt$$
Fantastic, I couldn't have come close. But as a general rule I wouldn't have used "t" as the integration variable in the second substitution. I know it's just a dummy variable but as we've already used it in the problem I find it to a point of potential confusion.

-Dan
 
  • #6
topsquark said:
Fantastic, I couldn't have come close. But as a general rule I wouldn't have used "t" as the integration variable in the second substitution. I know it's just a dummy variable but as we've already used it in the problem I find it to a point of potential confusion.

-Dan

Thanks. I edited my post , I was already thinking about that :)
 

FAQ: Solve Integral w/ Log: $$\int_{5}^{\infty}\frac{1}{t^r\log t}dt$$

What is an integral?

An integral is a mathematical concept that represents the area under a curve on a graph. It is used to calculate the total amount of a quantity, such as distance or area, by dividing it into infinitely small pieces and summing them up.

What is a logarithm?

A logarithm is the inverse function of an exponential, and it represents the power to which a base number must be raised to equal a given number. It is commonly used in mathematics to solve equations involving exponents and to express large numbers in a more manageable form.

How do you solve an integral with a logarithm?

To solve an integral with a logarithm, you can use the substitution method. In this case, let u = log t and du = 1/t dt. Then, the integral becomes ∫du/u^r. This can be solved using the power rule for integrals, giving the final answer as (-1/(r-1))u^(r-1). Don't forget to substitute back in for u to get the final answer.

What does the notation "∞" mean in an integral?

The symbol "∞" represents infinity, which in mathematics is a concept that refers to a quantity that has no upper bound. In an integral, it indicates that the upper limit of the integration is unbounded, or goes on forever.

How does the value of "r" affect the solution to this integral?

The value of "r" in this integral affects the convergence or divergence of the solution. If r is less than 1, the integral will converge and have a finite solution. If r is greater than or equal to 1, the integral will diverge and have an infinite solution. Additionally, the specific value of r will affect the exact value of the solution.

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