# Tricky Integrals

#### Phoenix314

All of these integrals have lower bounds of 0 and upper bounds of infinity:

Problems 1 and 2 just require me to determine whether it converges or diverges. 3 and 4 actually require a value.

1) e^(-x) * sqrt(x)
2) $$\frac{x*arctan(x)}{(1+x^4)^(1/3)}$$
3) e^(-x) / sqrt(x)
4) x^2 * e^[-(x^2)]

I tried to use integration by parts but I went in circles

[Edit:] number 2 should have (1+x^4)^(1/3) in the denominator

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#### lurflurf

Homework Helper
Phoenix314 said:
All of these integrals have lower bounds of 0 and upper bounds of infinity:

1) e^(-x) * sqrt(x)
2) $$\frac{x*arctan(x)}{(1+x^4)^(1/3)}$$
3) e^(-x) / sqrt(x)
4) x^2 * e^[-(x^2)]

I tried to use integration by parts but I went in circles

[Edit:] number 2 should have (1+x^4)^(1/3) in the denominator
Do you know the integral on that set for exp(-x^2)? 1,3,4 can be written in terms of it. If i am reading 2 right as (x*Arctan(x))/(1+x^4)^(1/3)~x^(-1/3) for large x and hence diverges.

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#### Phoenix314

I am given that e^(-x^2) = 0.5*sqrt(pi), but I couldn't get the integrals into that form through integration by parts - am I doing something wrong?

Thanks

#### lurflurf

Homework Helper
Phoenix314 said:
I am given that e^(-x^2) = 0.5*sqrt(pi), but I couldn't get the integrals into that form through integration by parts - am I doing something wrong?

Thanks
For 1,3 change variable u^2=x you will get 4 back from one of them and the given integral from the others. for 4 and the one that becomes like it intgrate by parts differentiate x and integrat x exp(-x^2). Again 2 (perhaps you mistyped it?) diverges.

#### dextercioby

Homework Helper
Are you sure about the second?

$$\int_{0}^{+\infty} \frac{x \arctan x}{\sqrt[3]{\left(1+x^{4}\right)}} \ dx$$

Maple cannot do it and neither Mathematica.

Daniel.

#### Phoenix314

That's the correct integral, it must diverge.

#### Phoenix314

These are the answers that I got

1) Converges (to 0.5*sqrt (Pi))
2) Diverges
3) Converges to sqrt (Pi)
4) Converges to 0.25 * sqrt (Pi)

Do these look right?

Thanks

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