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

[Bibubo]

How can I solve (or evaluating) this integral $$\int_{5}^{\infty}\frac{1}{\log\left(t\right)t^{r}}dt$$ with $r\geq 2$?

 

[Prove It]

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...

 

[alyafey22]

$$\int^\infty_5 \frac{1}{\log(t)\,t^r}\,dt$$

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

$$E_n(x) = \int^\infty_1 \frac{e^{-xt}}{t^n}\,dt$$

 

[Opalg]

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 $$\int_{(r-1)\log 5}^\infty \frac{e^{-x}}xdx$$, which cannot be expressed in terms of elementary functions.

Edit. Sorry, didn't see ZaidAlyafey's comment.

 

[topsquark]

$$\int^\infty_5 \frac{1}{\log(t)\,t^r}\,dt$$

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

 

[alyafey22]

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 :)